• 6.1 Exponential Functions
  • Introduction to Prerequisites
  • 1.1 Real Numbers: Algebra Essentials
  • 1.2 Exponents and Scientific Notation
  • 1.3 Radicals and Rational Exponents
  • 1.4 Polynomials
  • 1.5 Factoring Polynomials
  • 1.6 Rational Expressions
  • Key Equations
  • Key Concepts
  • Review Exercises
  • Practice Test
  • Introduction to Equations and Inequalities
  • 2.1 The Rectangular Coordinate Systems and Graphs
  • 2.2 Linear Equations in One Variable
  • 2.3 Models and Applications
  • 2.4 Complex Numbers
  • 2.5 Quadratic Equations
  • 2.6 Other Types of Equations
  • 2.7 Linear Inequalities and Absolute Value Inequalities
  • Introduction to Functions
  • 3.1 Functions and Function Notation
  • 3.2 Domain and Range
  • 3.3 Rates of Change and Behavior of Graphs
  • 3.4 Composition of Functions
  • 3.5 Transformation of Functions
  • 3.6 Absolute Value Functions
  • 3.7 Inverse Functions
  • Introduction to Linear Functions
  • 4.1 Linear Functions
  • 4.2 Modeling with Linear Functions
  • 4.3 Fitting Linear Models to Data
  • Introduction to Polynomial and Rational Functions
  • 5.1 Quadratic Functions
  • 5.2 Power Functions and Polynomial Functions
  • 5.3 Graphs of Polynomial Functions
  • 5.4 Dividing Polynomials
  • 5.5 Zeros of Polynomial Functions
  • 5.6 Rational Functions
  • 5.7 Inverses and Radical Functions
  • 5.8 Modeling Using Variation
  • Introduction to Exponential and Logarithmic Functions
  • 6.2 Graphs of Exponential Functions
  • 6.3 Logarithmic Functions
  • 6.4 Graphs of Logarithmic Functions
  • 6.5 Logarithmic Properties
  • 6.6 Exponential and Logarithmic Equations
  • 6.7 Exponential and Logarithmic Models
  • 6.8 Fitting Exponential Models to Data
  • Introduction to Systems of Equations and Inequalities
  • 7.1 Systems of Linear Equations: Two Variables
  • 7.2 Systems of Linear Equations: Three Variables
  • 7.3 Systems of Nonlinear Equations and Inequalities: Two Variables
  • 7.4 Partial Fractions
  • 7.5 Matrices and Matrix Operations
  • 7.6 Solving Systems with Gaussian Elimination
  • 7.7 Solving Systems with Inverses
  • 7.8 Solving Systems with Cramer's Rule
  • Introduction to Analytic Geometry
  • 8.1 The Ellipse
  • 8.2 The Hyperbola
  • 8.3 The Parabola
  • 8.4 Rotation of Axes
  • 8.5 Conic Sections in Polar Coordinates
  • Introduction to Sequences, Probability and Counting Theory
  • 9.1 Sequences and Their Notations
  • 9.2 Arithmetic Sequences
  • 9.3 Geometric Sequences
  • 9.4 Series and Their Notations
  • 9.5 Counting Principles
  • 9.6 Binomial Theorem
  • 9.7 Probability

Learning Objectives

In this section, you will:

  • Evaluate exponential functions.
  • Find the equation of an exponential function.
  • Use compound interest formulas.
  • Evaluate exponential functions with base e e .

India is the second most populous country in the world with a population of about 1.39 1.39 billion people in 2021. The population is growing at a rate of about 1.2 % 1.2 % each year 2 . If this rate continues, the population of India will exceed China’s population by the year 2027. 2027. When populations grow rapidly, we often say that the growth is “exponential,” meaning that something is growing very rapidly. To a mathematician, however, the term exponential growth has a very specific meaning. In this section, we will take a look at exponential functions , which model this kind of rapid growth.

Identifying Exponential Functions

When exploring linear growth, we observed a constant rate of change—a constant number by which the output increased for each unit increase in input. For example, in the equation f ( x ) = 3 x + 4 , f ( x ) = 3 x + 4 , the slope tells us the output increases by 3 each time the input increases by 1. The scenario in the India population example is different because we have a percent change per unit time (rather than a constant change) in the number of people.

Defining an Exponential Function

A study found that the percent of the population who are vegans in the United States doubled from 2009 to 2011. In 2011, 2.5% of the population was vegan, adhering to a diet that does not include any animal products—no meat, poultry, fish, dairy, or eggs. If this rate continues, vegans will make up 10% of the U.S. population in 2015, 40% in 2019, and 80% in 2021.

What exactly does it mean to grow exponentially ? What does the word double have in common with percent increase ? People toss these words around errantly. Are these words used correctly? The words certainly appear frequently in the media.

  • Percent change refers to a change based on a percent of the original amount.
  • Exponential growth refers to an increase based on a constant multiplicative rate of change over equal increments of time, that is, a percent increase of the original amount over time.
  • Exponential decay refers to a decrease based on a constant multiplicative rate of change over equal increments of time, that is, a percent decrease of the original amount over time.

For us to gain a clear understanding of exponential growth , let us contrast exponential growth with linear growth . We will construct two functions. The first function is exponential. We will start with an input of 0, and increase each input by 1. We will double the corresponding consecutive outputs. The second function is linear. We will start with an input of 0, and increase each input by 1. We will add 2 to the corresponding consecutive outputs. See Table 1 .

From Table 1 we can infer that for these two functions, exponential growth dwarfs linear growth.

  • Exponential growth refers to the original value from the range increases by the same percentage over equal increments found in the domain.
  • Linear growth refers to the original value from the range increases by the same amount over equal increments found in the domain.

Apparently, the difference between “the same percentage” and “the same amount” is quite significant. For exponential growth, over equal increments, the constant multiplicative rate of change resulted in doubling the output whenever the input increased by one. For linear growth, the constant additive rate of change over equal increments resulted in adding 2 to the output whenever the input was increased by one.

The general form of the exponential function is f ( x ) = a b x , f ( x ) = a b x , where a a is any nonzero number, b b is a positive real number not equal to 1.

  • If b > 1 , b > 1 , the function grows at a rate proportional to its size.
  • If 0 < b < 1 , 0 < b < 1 , the function decays at a rate proportional to its size.

Let’s look at the function f ( x ) = 2 x f ( x ) = 2 x from our example. We will create a table ( Table 2 ) to determine the corresponding outputs over an interval in the domain from − 3 − 3 to 3. 3.

Let us examine the graph of f f by plotting the ordered pairs we observe on the table in Figure 1 , and then make a few observations.

Let’s define the behavior of the graph of the exponential function f ( x ) = 2 x f ( x ) = 2 x and highlight some its key characteristics.

  • the domain is ( − ∞ , ∞ ) , ( − ∞ , ∞ ) ,
  • the range is ( 0 , ∞ ) , ( 0 , ∞ ) ,
  • as x → ∞ , f ( x ) → ∞ , x → ∞ , f ( x ) → ∞ ,
  • as x → − ∞ , f ( x ) → 0 , x → − ∞ , f ( x ) → 0 ,
  • f ( x ) f ( x ) is always increasing,
  • the graph of f ( x ) f ( x ) will never touch the x -axis because base two raised to any exponent never has the result of zero.
  • y = 0 y = 0 is the horizontal asymptote.
  • the y -intercept is 1.

Exponential Function

For any real number x , x , an exponential function is a function with the form

  • a a is a non-zero real number called the initial value and
  • b b is any positive real number such that b ≠ 1. b ≠ 1.
  • The domain of f f is all real numbers.
  • The range of f f is all positive real numbers if a > 0. a > 0.
  • The range of f f is all negative real numbers if a < 0. a < 0.
  • The y -intercept is ( 0 , a ) , ( 0 , a ) , and the horizontal asymptote is y = 0. y = 0.

Which of the following equations are not exponential functions?

  • f ( x ) = 4 3 ( x − 2 ) f ( x ) = 4 3 ( x − 2 )
  • g ( x ) = x 3 g ( x ) = x 3
  • h ( x ) = ( 1 3 ) x h ( x ) = ( 1 3 ) x
  • j ( x ) = ( − 2 ) x j ( x ) = ( − 2 ) x

By definition, an exponential function has a constant as a base and an independent variable as an exponent. Thus, g ( x ) = x 3 g ( x ) = x 3 does not represent an exponential function because the base is an independent variable. In fact, g ( x ) = x 3 g ( x ) = x 3 is a power function.

Recall that the base b of an exponential function is always a positive constant, and b ≠ 1. b ≠ 1. Thus, j ( x ) = ( −2 ) x j ( x ) = ( −2 ) x does not represent an exponential function because the base, −2 , −2 , is less than 0. 0.

Which of the following equations represent exponential functions?

  • f ( x ) = 2 x 2 − 3 x + 1 f ( x ) = 2 x 2 − 3 x + 1
  • g ( x ) = 0.875 x g ( x ) = 0.875 x
  • h ( x ) = 1.75 x + 2 h ( x ) = 1.75 x + 2
  • j ( x ) = 1095.6 − 2 x j ( x ) = 1095.6 − 2 x

Evaluating Exponential Functions

Recall that the base of an exponential function must be a positive real number other than 1. 1. Why do we limit the base b b to positive values? To ensure that the outputs will be real numbers. Observe what happens if the base is not positive:

  • Let b = − 9 b = − 9 and x = 1 2 . x = 1 2 . Then f ( x ) = f ( 1 2 ) = ( − 9 ) 1 2 = − 9 , f ( x ) = f ( 1 2 ) = ( − 9 ) 1 2 = − 9 , which is not a real number.

Why do we limit the base to positive values other than 1 ? 1 ? Because base 1 1 results in the constant function. Observe what happens if the base is 1 : 1 :

  • Let b = 1. b = 1. Then f ( x ) = 1 x = 1 f ( x ) = 1 x = 1 for any value of x . x .

To evaluate an exponential function with the form f ( x ) = b x , f ( x ) = b x , we simply substitute x x with the given value, and calculate the resulting power. For example:

Let f ( x ) = 2 x . f ( x ) = 2 x . What is f ( 3 ) ? f ( 3 ) ?

To evaluate an exponential function with a form other than the basic form, it is important to follow the order of operations. For example:

Let f ( x ) = 30 ( 2 ) x . f ( x ) = 30 ( 2 ) x . What is f ( 3 ) ? f ( 3 ) ?

Note that if the order of operations were not followed, the result would be incorrect:

Let f ( x ) = 5 ( 3 ) x + 1 . f ( x ) = 5 ( 3 ) x + 1 . Evaluate f ( 2 ) f ( 2 ) without using a calculator.

Follow the order of operations. Be sure to pay attention to the parentheses.

Let f ( x ) = 8 ( 1.2 ) x − 5 . f ( x ) = 8 ( 1.2 ) x − 5 . Evaluate f ( 3 ) f ( 3 ) using a calculator. Round to four decimal places.

Defining Exponential Growth

Because the output of exponential functions increases very rapidly, the term “exponential growth” is often used in everyday language to describe anything that grows or increases rapidly. However, exponential growth can be defined more precisely in a mathematical sense. If the growth rate is proportional to the amount present, the function models exponential growth.

Exponential Growth

A function that models exponential growth grows by a rate proportional to the amount present. For any real number x x and any positive real numbers a   a   and b b such that b ≠ 1 , b ≠ 1 , an exponential growth function has the form

  • a a is the initial or starting value of the function.
  • b b is the growth factor or growth multiplier per unit x x .

In more general terms, we have an exponential function , in which a constant base is raised to a variable exponent. To differentiate between linear and exponential functions, let’s consider two companies, A and B. Company A has 100 stores and expands by opening 50 new stores a year, so its growth can be represented by the function A ( x ) = 100 + 50 x . A ( x ) = 100 + 50 x . Company B has 100 stores and expands by increasing the number of stores by 50% each year, so its growth can be represented by the function B ( x ) = 100 ( 1 + 0.5 ) x . B ( x ) = 100 ( 1 + 0.5 ) x .

A few years of growth for these companies are illustrated in Table 3 .

The graphs comparing the number of stores for each company over a five-year period are shown in Figure 2 . We can see that, with exponential growth, the number of stores increases much more rapidly than with linear growth.

Notice that the domain for both functions is [ 0 , ∞ ) , [ 0 , ∞ ) , and the range for both functions is [ 100 , ∞ ) . [ 100 , ∞ ) . After year 1, Company B always has more stores than Company A.

Now we will turn our attention to the function representing the number of stores for Company B, B ( x ) = 100 ( 1 + 0.5 ) x . B ( x ) = 100 ( 1 + 0.5 ) x . In this exponential function, 100 represents the initial number of stores, 0.50 represents the growth rate, and 1 + 0.5 = 1.5 1 + 0.5 = 1.5 represents the growth factor. Generalizing further, we can write this function as B ( x ) = 100 ( 1.5 ) x , B ( x ) = 100 ( 1.5 ) x , where 100 is the initial value, 1.5 1.5 is called the base , and x x is called the exponent .

Evaluating a Real-World Exponential Model

At the beginning of this section, we learned that the population of India was about 1.25 1.25 billion in the year 2013, with an annual growth rate of about 1.2 % . 1.2 % . This situation is represented by the growth function P ( t ) = 1.25 ( 1.012 ) t , P ( t ) = 1.25 ( 1.012 ) t , where t t is the number of years since 2013. 2013. To the nearest thousandth, what will the population of India be in 2031? 2031?

To estimate the population in 2031, we evaluate the models for t = 18 , t = 18 , because 2031 is 18 18 years after 2013. Rounding to the nearest thousandth,

There will be about 1.549 billion people in India in the year 2031.

The population of China was about 1.39 billion in the year 2013, with an annual growth rate of about 0.6 % . 0.6 % . This situation is represented by the growth function P ( t ) = 1.39 ( 1.006 ) t , P ( t ) = 1.39 ( 1.006 ) t , where t t is the number of years since 2013. 2013. To the nearest thousandth, what will the population of China be for the year 2031? How does this compare to the population prediction we made for India in Example 3 ?

Finding Equations of Exponential Functions

In the previous examples, we were given an exponential function, which we then evaluated for a given input. Sometimes we are given information about an exponential function without knowing the function explicitly. We must use the information to first write the form of the function, then determine the constants a a and b , b , and evaluate the function.

Given two data points, write an exponential model.

  • If one of the data points has the form ( 0 , a ) , ( 0 , a ) , then a a is the initial value. Using a , a , substitute the second point into the equation f ( x ) = a ( b ) x , f ( x ) = a ( b ) x , and solve for b . b .
  • If neither of the data points have the form ( 0 , a ) , ( 0 , a ) , substitute both points into two equations with the form f ( x ) = a ( b ) x . f ( x ) = a ( b ) x . Solve the resulting system of two equations in two unknowns to find a a and b . b .
  • Using the a a and b b found in the steps above, write the exponential function in the form f ( x ) = a ( b ) x . f ( x ) = a ( b ) x .

Writing an Exponential Model When the Initial Value Is Known

In 2006, 80 deer were introduced into a wildlife refuge. By 2012, the population had grown to 180 deer. The population was growing exponentially. Write an exponential function N ( t ) N ( t ) representing the population ( N ) ( N ) of deer over time t . t .

We let our independent variable t t be the number of years after 2006. Thus, the information given in the problem can be written as input-output pairs: (0, 80) and (6, 180). Notice that by choosing our input variable to be measured as years after 2006, we have given ourselves the initial value for the function, a = 80. a = 80. We can now substitute the second point into the equation N ( t ) = 80 b t N ( t ) = 80 b t to find b : b :

NOTE: Unless otherwise stated, do not round any intermediate calculations. Then round the final answer to four places for the remainder of this section.

The exponential model for the population of deer is N ( t ) = 80 ( 1.1447 ) t . N ( t ) = 80 ( 1.1447 ) t . (Note that this exponential function models short-term growth. As the inputs gets large, the output will get increasingly larger, so much so that the model may not be useful in the long term.)

We can graph our model to observe the population growth of deer in the refuge over time. Notice that the graph in Figure 3 passes through the initial points given in the problem, ( 0 , 80 ) ( 0 , 80 ) and ( 6 , 180 ) . ( 6 , 180 ) . We can also see that the domain for the function is [ 0 , ∞ ) , [ 0 , ∞ ) , and the range for the function is [ 80 , ∞ ) . [ 80 , ∞ ) .

A wolf population is growing exponentially. In 2011, 129 129 wolves were counted. By 2013, 2013, the population had reached 236 wolves. What two points can be used to derive an exponential equation modeling this situation? Write the equation representing the population N N of wolves over time t . t .

Writing an Exponential Model When the Initial Value is Not Known

Find an exponential function that passes through the points ( − 2 , 6 ) ( − 2 , 6 ) and ( 2 , 1 ) . ( 2 , 1 ) .

Because we don’t have the initial value, we substitute both points into an equation of the form f ( x ) = a b x , f ( x ) = a b x , and then solve the system for a a and b . b .

  • Substituting ( − 2 , 6 ) ( − 2 , 6 ) gives 6 = a b − 2 6 = a b − 2
  • Substituting ( 2 , 1 ) ( 2 , 1 ) gives 1 = a b 2 1 = a b 2

Use the first equation to solve for a a in terms of b : b :

Substitute a a in the second equation, and solve for b : b :

Use the value of b b in the first equation to solve for the value of a : a :

Thus, the equation is f ( x ) = 2.4492 ( 0.6389 ) x . f ( x ) = 2.4492 ( 0.6389 ) x .

We can graph our model to check our work. Notice that the graph in Figure 4 passes through the initial points given in the problem, ( − 2 , 6 ) ( − 2 , 6 ) and ( 2 , 1 ) . ( 2 , 1 ) . The graph is an example of an exponential decay function.

Given the two points ( 1 , 3 ) ( 1 , 3 ) and ( 2 , 4.5 ) , ( 2 , 4.5 ) , find the equation of the exponential function that passes through these two points.

Do two points always determine a unique exponential function?

Yes, provided the two points are either both above the x-axis or both below the x-axis and have different x-coordinates. But keep in mind that we also need to know that the graph is, in fact, an exponential function. Not every graph that looks exponential really is exponential. We need to know the graph is based on a model that shows the same percent growth with each unit increase in x , x , which in many real world cases involves time.

Given the graph of an exponential function, write its equation.

  • First, identify two points on the graph. Choose the y -intercept as one of the two points whenever possible. Try to choose points that are as far apart as possible to reduce round-off error.
  • If one of the data points is the y- intercept ( 0 , a ) ( 0 , a ) , then a a is the initial value. Using a , a , substitute the second point into the equation f ( x ) = a ( b ) x , f ( x ) = a ( b ) x , and solve for b . b .
  • Write the exponential function, f ( x ) = a ( b ) x . f ( x ) = a ( b ) x .

Writing an Exponential Function Given Its Graph

Find an equation for the exponential function graphed in Figure 5 .

We can choose the y -intercept of the graph, ( 0 , 3 ) , ( 0 , 3 ) , as our first point. This gives us the initial value, a = 3. a = 3. Next, choose a point on the curve some distance away from ( 0 , 3 ) ( 0 , 3 ) that has integer coordinates. One such point is ( 2 , 12 ) . ( 2 , 12 ) .

Because we restrict ourselves to positive values of b , b , we will use b = 2. b = 2. Substitute a a and b b into the standard form to yield the equation f ( x ) = 3 ( 2 ) x . f ( x ) = 3 ( 2 ) x .

Find an equation for the exponential function graphed in Figure 6 .

Given two points on the curve of an exponential function, use a graphing calculator to find the equation.

  • Press [STAT].
  • Clear any existing entries in columns L1 or L2.
  • In L1 , enter the x -coordinates given.
  • In L2 , enter the corresponding y -coordinates.
  • Press [STAT] again. Cursor right to CALC , scroll down to ExpReg (Exponential Regression) , and press [ENTER].
  • The screen displays the values of a and b in the exponential equation y = a ⋅ b x y = a ⋅ b x .

Using a Graphing Calculator to Find an Exponential Function

Use a graphing calculator to find the exponential equation that includes the points ( 2 , 24.8 ) ( 2 , 24.8 ) and ( 5 , 198.4 ) . ( 5 , 198.4 ) .

Follow the guidelines above. First press [STAT] , [EDIT] , [1: Edit…], and clear the lists L1 and L2 . Next, in the L1 column, enter the x -coordinates, 2 and 5. Do the same in the L2 column for the y -coordinates, 24.8 and 198.4.

Now press [STAT] , [CALC] , [0: ExpReg] and press [ENTER] . The values a = 6.2 a = 6.2 and b = 2 b = 2 will be displayed. The exponential equation is y = 6.2 ⋅ 2 x . y = 6.2 ⋅ 2 x .

Use a graphing calculator to find the exponential equation that includes the points (3, 75.98) and (6, 481.07).

Applying the Compound-Interest Formula

Savings instruments in which earnings are continually reinvested, such as mutual funds and retirement accounts, use compound interest . The term compounding refers to interest earned not only on the original value, but on the accumulated value of the account.

The annual percentage rate (APR) of an account, also called the nominal rate , is the yearly interest rate earned by an investment account. The term  nominal  is used when the compounding occurs a number of times other than once per year. In fact, when interest is compounded more than once a year, the effective interest rate ends up being greater than the nominal rate! This is a powerful tool for investing.

We can calculate the compound interest using the compound interest formula, which is an exponential function of the variables time t , t , principal P , P , APR r , r , and number of compounding periods in a year n : n :

For example, observe Table 4 , which shows the result of investing $1,000 at 10% for one year. Notice how the value of the account increases as the compounding frequency increases.

The Compound Interest Formula

Compound interest can be calculated using the formula

  • A ( t ) A ( t ) is the account value,
  • t t is measured in years,
  • P P is the starting amount of the account, often called the principal, or more generally present value,
  • r r is the annual percentage rate (APR) expressed as a decimal, and
  • n n is the number of compounding periods in one year.

Calculating Compound Interest

If we invest $3,000 in an investment account paying 3% interest compounded quarterly, how much will the account be worth in 10 years?

Because we are starting with $3,000, P = 3000. P = 3000. Our interest rate is 3%, so r = 0.03. r = 0.03. Because we are compounding quarterly, we are compounding 4 times per year, so n = 4. n = 4. We want to know the value of the account in 10 years, so we are looking for A ( 10 ) , A ( 10 ) , the value when t = 10. t = 10.

The account will be worth about $4,045.05 in 10 years.

An initial investment of $100,000 at 12% interest is compounded weekly (use 52 weeks in a year). What will the investment be worth in 30 years?

Using the Compound Interest Formula to Solve for the Principal

A 529 Plan is a college-savings plan that allows relatives to invest money to pay for a child’s future college tuition; the account grows tax-free. Lily wants to set up a 529 account for her new granddaughter and wants the account to grow to $40,000 over 18 years. She believes the account will earn 6% compounded semi-annually (twice a year). To the nearest dollar, how much will Lily need to invest in the account now?

The nominal interest rate is 6%, so r = 0.06. r = 0.06. Interest is compounded twice a year, so n = 2. n = 2.

We want to find the initial investment, P , P , needed so that the value of the account will be worth $40,000 in 18 18 years. Substitute the given values into the compound interest formula, and solve for P . P .

Lily will need to invest $13,801 to have $40,000 in 18 years.

Refer to Example 9 . To the nearest dollar, how much would Lily need to invest if the account is compounded quarterly?

Evaluating Functions with Base e

As we saw earlier, the amount earned on an account increases as the compounding frequency increases. Table 5 shows that the increase from annual to semi-annual compounding is larger than the increase from monthly to daily compounding. This might lead us to ask whether this pattern will continue.

Examine the value of $1 invested at 100% interest for 1 year, compounded at various frequencies, listed in Table 5 .

These values appear to be approaching a limit as n n increases without bound. In fact, as n n gets larger and larger, the expression ( 1 + 1 n ) n ( 1 + 1 n ) n approaches a number used so frequently in mathematics that it has its own name: the letter e . e . This value is an irrational number, which means that its decimal expansion goes on forever without repeating. Its approximation to six decimal places is shown below.

The Number e e

The letter e represents the irrational number

The letter e is used as a base for many real-world exponential models. To work with base e , we use the approximation, e ≈ 2.718282. e ≈ 2.718282. The constant was named by the Swiss mathematician Leonhard Euler (1707–1783) who first investigated and discovered many of its properties.

Using a Calculator to Find Powers of e

Calculate e 3.14 . e 3.14 . Round to five decimal places.

On a calculator, press the button labeled [ e x ] . [ e x ] . The window shows [ e ^ ( ] . [ e ^ ( ] . Type 3.14 3.14 and then close parenthesis, [ ) ] . [ ) ] . Press [ENTER]. Rounding to 5 5 decimal places, e 3.14 ≈ 23.10387. e 3.14 ≈ 23.10387. Caution: Many scientific calculators have an “Exp” button, which is used to enter numbers in scientific notation. It is not used to find powers of e . e .

Use a calculator to find e − 0.5 . e − 0.5 . Round to five decimal places.

Investigating Continuous Growth

So far we have worked with rational bases for exponential functions. For most real-world phenomena, however, e is used as the base for exponential functions. Exponential models that use e e as the base are called continuous growth or decay models . We see these models in finance, computer science, and most of the sciences, such as physics, toxicology, and fluid dynamics.

The Continuous Growth/Decay Formula

For all real numbers t , t , and all positive numbers a a and r , r , continuous growth or decay is represented by the formula

  • a a is the initial value,
  • r r is the continuous growth rate per unit time,
  • and t t is the elapsed time.

If r > 0 r > 0 , then the formula represents continuous growth. If r < 0 r < 0 , then the formula represents continuous decay.

For business applications, the continuous growth formula is called the continuous compounding formula and takes the form

  • P P is the principal or the initial invested,
  • r r is the growth or interest rate per unit time,
  • and t t is the period or term of the investment.

Given the initial value, rate of growth or decay, and time t , t , solve a continuous growth or decay function.

  • Use the information in the problem to determine a a , the initial value of the function.
  • If the problem refers to continuous growth, then r > 0. r > 0.
  • If the problem refers to continuous decay, then r < 0. r < 0.
  • Use the information in the problem to determine the time t . t .
  • Substitute the given information into the continuous growth formula and solve for A ( t ) . A ( t ) .

Calculating Continuous Growth

A person invested $1,000 in an account earning a nominal 10% per year compounded continuously. How much was in the account at the end of one year?

Since the account is growing in value, this is a continuous compounding problem with growth rate r = 0.10. r = 0.10. The initial investment was $1,000, so P = 1000. P = 1000. We use the continuous compounding formula to find the value after t = 1 t = 1 year:

The account is worth $1,105.17 after one year.

A person invests $100,000 at a nominal 12% interest per year compounded continuously. What will be the value of the investment in 30 years?

Calculating Continuous Decay

Radon-222 decays at a continuous rate of 17.3% per day. How much will 100 mg of Radon-222 decay to in 3 days?

Since the substance is decaying, the rate, 17.3 % 17.3 % , is negative. So, r = − 0.173. r = − 0.173. The initial amount of radon-222 was 100 100 mg, so a = 100. a = 100. We use the continuous decay formula to find the value after t = 3 t = 3 days:

So 59.5115 mg of radon-222 will remain.

Using the data in Example 12 , how much radon-222 will remain after one year?

Access these online resources for additional instruction and practice with exponential functions.

  • Exponential Growth Function
  • Compound Interest

Explain why the values of an increasing exponential function will eventually overtake the values of an increasing linear function.

Given a formula for an exponential function, is it possible to determine whether the function grows or decays exponentially just by looking at the formula? Explain.

The Oxford Dictionary defines the word nominal as a value that is “stated or expressed but not necessarily corresponding exactly to the real value.” 3 Develop a reasonable argument for why the term nominal rate is used to describe the annual percentage rate of an investment account that compounds interest.

For the following exercises, identify whether the statement represents an exponential function. Explain.

The average annual population increase of a pack of wolves is 25.

A population of bacteria decreases by a factor of 1 8 1 8 every 24 24 hours.

The value of a coin collection has increased by 3.25 % 3.25 % annually over the last 20 20 years.

For each training session, a personal trainer charges his clients $ 5 $ 5 less than the previous training session.

The height of a projectile at time t t is represented by the function h ( t ) = − 4.9 t 2 + 18 t + 40. h ( t ) = − 4.9 t 2 + 18 t + 40.

For the following exercises, consider this scenario: For each year t , t , the population of a forest of trees is represented by the function A ( t ) = 115 ( 1.025 ) t . A ( t ) = 115 ( 1.025 ) t . In a neighboring forest, the population of the same type of tree is represented by the function B ( t ) = 82 ( 1.029 ) t . B ( t ) = 82 ( 1.029 ) t . (Round answers to the nearest whole number.)

Which forest’s population is growing at a faster rate?

Which forest had a greater number of trees initially? By how many?

Assuming the population growth models continue to represent the growth of the forests, which forest will have a greater number of trees after 20 20 years? By how many?

Assuming the population growth models continue to represent the growth of the forests, which forest will have a greater number of trees after 100 100 years? By how many?

Discuss the above results from the previous four exercises. Assuming the population growth models continue to represent the growth of the forests, which forest will have the greater number of trees in the long run? Why? What are some factors that might influence the long-term validity of the exponential growth model?

For the following exercises, determine whether the equation represents exponential growth, exponential decay, or neither. Explain.

y = 300 ( 1 − t ) 5 y = 300 ( 1 − t ) 5

y = 220 ( 1.06 ) x y = 220 ( 1.06 ) x

y = 16.5 ( 1.025 ) 1 x y = 16.5 ( 1.025 ) 1 x

y = 11 , 701 ( 0.97 ) t y = 11 , 701 ( 0.97 ) t

For the following exercises, find the formula for an exponential function that passes through the two points given.

( 0 , 6 ) ( 0 , 6 ) and ( 3 , 750 ) ( 3 , 750 )

( 0 , 2000 ) ( 0 , 2000 ) and ( 2 , 20 ) ( 2 , 20 )

( − 1 , 3 2 ) ( − 1 , 3 2 ) and ( 3 , 24 ) ( 3 , 24 )

( − 2 , 6 ) ( − 2 , 6 ) and ( 3 , 1 ) ( 3 , 1 )

( 3 , 1 ) ( 3 , 1 ) and ( 5 , 4 ) ( 5 , 4 )

For the following exercises, determine whether the table could represent a function that is linear, exponential, or neither. If it appears to be exponential, find a function that passes through the points.

For the following exercises, use the compound interest formula, A ( t ) = P ( 1 + r n ) n t . A ( t ) = P ( 1 + r n ) n t .

After a certain number of years, the value of an investment account is represented by the equation A = 10 , 250 ( 1 + 0.04 12 ) 120 . A = 10 , 250 ( 1 + 0.04 12 ) 120 . What is the value of the account?

What was the initial deposit made to the account in the previous exercise?

How many years had the account from the previous exercise been accumulating interest?

An account is opened with an initial deposit of $6,500 and earns 3.6 % 3.6 % interest compounded semi-annually. What will the account be worth in 20 20 years?

How much more would the account in the previous exercise have been worth if the interest were compounding weekly?

Solve the compound interest formula for the principal, P P .

Use the formula found in the previous exercise to calculate the initial deposit of an account that is worth $ 14 , 472.74 $ 14 , 472.74 after earning 5.5 % 5.5 % interest compounded monthly for 5 5 years. (Round to the nearest dollar.)

How much more would the account in the previous two exercises be worth if it were earning interest for 5 5 more years?

Use properties of rational exponents to solve the compound interest formula for the interest rate, r . r .

Use the formula found in the previous exercise to calculate the interest rate for an account that was compounded semi-annually, had an initial deposit of $9,000 and was worth $13,373.53 after 10 years.

Use the formula found in the previous exercise to calculate the interest rate for an account that was compounded monthly, had an initial deposit of $5,500, and was worth $38,455 after 30 years.

For the following exercises, determine whether the equation represents continuous growth, continuous decay, or neither. Explain.

y = 3742 ( e ) 0.75 t y = 3742 ( e ) 0.75 t

y = 150 ( e ) 3.25 t y = 150 ( e ) 3.25 t

y = 2.25 ( e ) − 2 t y = 2.25 ( e ) − 2 t

Suppose an investment account is opened with an initial deposit of $ 12 , 000 $ 12 , 000 earning 7.2 % 7.2 % interest compounded continuously. How much will the account be worth after 30 30 years?

How much less would the account from Exercise 42 be worth after 30 30 years if it were compounded monthly instead?

For the following exercises, evaluate each function. Round answers to four decimal places, if necessary.

f ( x ) = 2 ( 5 ) x , f ( x ) = 2 ( 5 ) x , for f ( − 3 ) f ( − 3 )

f ( x ) = − 4 2 x + 3 , f ( x ) = − 4 2 x + 3 , for f ( − 1 ) f ( − 1 )

f ( x ) = e x , f ( x ) = e x , for f ( 3 ) f ( 3 )

f ( x ) = − 2 e x − 1 , f ( x ) = − 2 e x − 1 , for f ( − 1 ) f ( − 1 )

f ( x ) = 2.7 ( 4 ) − x + 1 + 1.5 , f ( x ) = 2.7 ( 4 ) − x + 1 + 1.5 , for f ( − 2 ) f ( − 2 )

f ( x ) = 1.2 e 2 x − 0.3 , f ( x ) = 1.2 e 2 x − 0.3 , for f ( 3 ) f ( 3 )

f ( x ) = − 3 2 ( 3 ) − x + 3 2 , f ( x ) = − 3 2 ( 3 ) − x + 3 2 , for f ( 2 ) f ( 2 )

For the following exercises, use a graphing calculator to find the equation of an exponential function given the points on the curve.

( 0 , 3 ) ( 0 , 3 ) and ( 3 , 375 ) ( 3 , 375 )

( 3 , 222.62 ) ( 3 , 222.62 ) and ( 10 , 77.456 ) ( 10 , 77.456 )

( 20 , 29.495 ) ( 20 , 29.495 ) and ( 150 , 730.89 ) ( 150 , 730.89 )

( 5 , 2.909 ) ( 5 , 2.909 ) and ( 13 , 0.005 ) ( 13 , 0.005 )

( 11,310.035 ) ( 11,310.035 ) and ( 25,356365.2 ) ( 25,356365.2 )

The annual percentage yield (APY) of an investment account is a representation of the actual interest rate earned on a compounding account. It is based on a compounding period of one year. Show that the APY of an account that compounds monthly can be found with the formula APY = ( 1 + r 12 ) 12 − 1. APY = ( 1 + r 12 ) 12 − 1.

Repeat the previous exercise to find the formula for the APY of an account that compounds daily. Use the results from this and the previous exercise to develop a function I ( n ) I ( n ) for the APY of any account that compounds n n times per year.

Recall that an exponential function is any equation written in the form f ( x ) = a ⋅ b x f ( x ) = a ⋅ b x such that   a     a   and   b     b   are positive numbers and   b ≠ 1.     b ≠ 1.   Any positive number   b     b   can be written as   b = e n     b = e n   for some value of   n   n . Use this fact to rewrite the formula for an exponential function that uses the number   e     e   as a base.

In an exponential decay function, the base of the exponent is a value between 0 and 1. Thus, for some number b > 1 , b > 1 , the exponential decay function can be written as f ( x ) = a ⋅ ( 1 b ) x . f ( x ) = a ⋅ ( 1 b ) x . Use this formula, along with the fact that b = e n , b = e n , to show that an exponential decay function takes the form f ( x ) = a ( e ) − n x f ( x ) = a ( e ) − n x for some positive number n n .

The formula for the amount A A in an investment account with a nominal interest rate r r at any time t t is given by A ( t ) = a ( e ) r t , A ( t ) = a ( e ) r t , where a a is the amount of principal initially deposited into an account that compounds continuously. Prove that the percentage of interest earned to principal at any time t t can be calculated with the formula I ( t ) = e r t − 1. I ( t ) = e r t − 1.

Real-World Applications

The fox population in a certain region has an annual growth rate of 9% per year. In the year 2012, there were 23,900 fox counted in the area. What is the fox population predicted to be in the year 2020?

A scientist begins with 100 milligrams of a radioactive substance that decays exponentially. After 35 hours, 50mg of the substance remains. How many milligrams will remain after 54 hours?

In the year 1985, a house was valued at $110,000. By the year 2005, the value had appreciated to $145,000. What was the annual growth rate between 1985 and 2005? Assume that the value continued to grow by the same percentage. What was the value of the house in the year 2010?

A car was valued at $38,000 in the year 2007. By 2013, the value had depreciated to $11,000 If the car’s value continues to drop by the same percentage, what will it be worth by 2017?

Jaylen wants to save $54,000 for a down payment on a home. How much will he need to invest in an account with 8.2% APR, compounding daily, in order to reach his goal in 5 years?

Kyoko has $10,000 that she wants to invest. Her bank has several investment accounts to choose from, all compounding daily. Her goal is to have $15,000 by the time she finishes graduate school in 6 years. To the nearest hundredth of a percent, what should her minimum annual interest rate be in order to reach her goal? ( Hint : solve the compound interest formula for the interest rate.)

Alyssa opened a retirement account with 7.25% APR in the year 2000. Her initial deposit was $13,500. How much will the account be worth in 2025 if interest compounds monthly? How much more would she make if interest compounded continuously?

An investment account with an annual interest rate of 7% was opened with an initial deposit of $4,000 Compare the values of the account after 9 years when the interest is compounded annually, quarterly, monthly, and continuously.

  • 2 http://www.worldometers.info/world-population/. Accessed February 24, 2014.
  • 3 Oxford Dictionary. http://oxforddictionaries.com/us/definition/american_english/nomina.

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  • Book title: College Algebra 2e
  • Publication date: Dec 21, 2021
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  • Book URL: https://openstax.org/books/college-algebra-2e/pages/1-introduction-to-prerequisites
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  • Applications of Exponential Functions

Key Questions

John tells you a secret. You see no harm in telling Bob and Rob .

After this, 4 people know the secret (John, you Bob and Rob). Suppose that both Bob and Rob decide to tell the secret to two new people.

After the third "round" of indiscretion, eight people will know the secret.

If this pattern of spreading the secret continues, how many people will know the secret after 10 such "rounds"?

homework 10 applications of exponential functions

World-Population from 1800 (actual) to 2100 (projected): The projected world population growths after the present day must be projected. A high estimation predicts the graph to continue at an increasing rate, a medium estimation predicts the population to level off, and a low estimation predicts the population to decline.

The graph has the general shape of an exponential curve though it is not exact as is the case usually when we deal with real data as opposed to purely mathematical constructs. The data is at times estimated, at times actual and at times projected. Projections for high (red), medium (orange) and low (green) rates of change are represented accounting for the splitting off of the curve after 2016.

Limited Growth

A realistic model of exponential growth must dampen when approaching a certain value. This limited growth is modelled with the logistic growth model: [latex]P(t)=\frac{c}{1+a\cdot e^{-bt}}[/latex].

Use logistic functions to represent limited growth

  • Exponential growth may exist within known parameters, but such a functionality may not continue indefinitely in the real world.
  • If a natural maximum is conceivable, the logistic growth model can be used to represent growth. It has the form: [latex]P(t)=\frac{c}{1+a\cdot e^{-bt}}[/latex]where [latex]P[/latex] represents population, [latex]c[/latex] is the carrying capacity, [latex]b[/latex] is the population growth rate, [latex]t[/latex] is time, and [latex]a[/latex] is the difference between carrying capacity and initial population.
  • An example of natural dampening in growth is the population of humans on planet Earth. The population may be growing exponentially at the moment, but eventually, scarcity of resources will curb our growth as we reach our carrying capacity.
  • logistic function : A defined function that is the result of the division of two exponential functions. When plotted it gives the logistic curve.

Exponential functions can be used to model growth and decay. For example, the world’s human population is growing exponentially as can be seen in the following graph.

image

World Population from 1800 AD to 2100 AD: Three projections for the world’s population are shown, with a low estimate reaching a peak and then decreasing, a medium estimate increasing but at an ever-slower rate, and a high estimate continuing to increase exponentially.

There is, however, a limit to how accurate a model such as this can be. Even if we account for varying rates of growth, the idea that human population can be modeled strictly with an exponential function is misguided. Exponential functions are ever-increasing so saying that an exponential function models population growth exactly means that the human population will grow without bound. That is not sound reasoning, as the human population is affected by various factors among these are access to resources such as food, water, and shelter. At some point in the future, the number of humans will grow so large that there will not be enough resources to sustain growth. Thus the exponential model, while perhaps useful in the short-term, is not accurate in the long-term for real world simulation. At some point the growth must level off. This is the idea behind limited growth, that a population may grow exponentially but only to a point at which time the growth will taper off.

As an alternative, consider a farm upon which a population of sheep are kept in a constant, comfortable climate in a fully enclosed field. Assume the entire system is closed from gains and losses, but for a flow of a stream of clean drinking water through the field.

With no predators, the population of sheep would perfectly fit an exponential model to a certain point when the availability of food would act as a limiting factor. If the amount of grass available to the sheep and its rate of replenishing are constant, eventually the population of sheep will grow to a tipping point at which there is no longer enough grass to feed the sheep. The death rate of sheep will increase as some starve. Thus, the model of population growth among sheep will no longer be exponential.

Logistic Growth Model

To account for limitations in growth, the logistic growth model can be used. The logistic growth model is given by [latex]P(t)=\frac{c}{1+a\cdot e^{-bt}}[/latex] where [latex]P[/latex] represents the present population, [latex]c[/latex] is the carrying capacity (the maximum the population approaches as time approaches infinity), [latex]b[/latex] is the population growth rate, [latex]t[/latex] is time, and [latex]a[/latex] is the difference between carrying capacity and initial population.

Evaluating a Logistic Growth Function

Given various conditions, it is possible to evaluate a logistic function for a particular value of [latex]t[/latex]. That is, it is possible to determine the population at time [latex]t[/latex], given values for [latex]c,a,b[/latex] and [latex]t[/latex].

Example 1: Evaluate the logistic growth function [latex]P(t)=\frac{50}{1+8\cdot e^{-3t}}[/latex] for [latex]t=6[/latex]

To evaluate this function we plug [latex]6[/latex] into the equation in place of [latex]t[/latex] as follows.

[latex]\displaystyle \begin{align} P(6)&=\frac{50}{1+8\cdot e^{(-3)(6)}}\\& \approx49.99999391 \end{align}[/latex]

Graphing a Logistic Growth Model

Graphically, the logistic function resembles an exponential function followed by a logarithmic function that approaches a horizontal asymptote. This horizontal asymptote represents the carrying capacity. That is, [latex]y=c[/latex] is a horizontal asymptote of the graph. Additionally, [latex]y=o[/latex] is also a horizontal asymptote. From the left, it grows rapidly, but that growth is dampened as time passes to where it reaches a maximum. The function’s domain is the set of all real numbers, whereas its range is [latex]0<y<c[/latex]. Below is the graph of a logistic function.

image

Graph of a logistic function: Logistic functions have an “s” shape, where the function starts from a certain point, increases, and then approaches an upper asymptote.

Interest Compounded Continuously

Compound interest is accrued when interest is earned not only on principal, but on previously accrued interest: it is interest on interest.

Differentiate between simple interest and compound interest

  • Interest is, generally, a fee charged for the borrowing of money. The amount of interest accrued depends on the principal (amount borrowed), the interest rate (a percentage of the principal), period (amount of time between interest payments) and time elapsed.
  • The form of the equation for compound interest is exponential, and thus such interest is accrued much faster than the linear simple interest.
  • The formula for compound interest is [latex]M=p(1+r)^\frac{t}{f}[/latex]Where [latex]M[/latex] represents the total value (including principal), [latex]p[/latex] represents principal, [latex]r[/latex] is interest rate (expressed as a decimal), [latex]t[/latex] is time elapsed, and [latex]f[/latex] is the length of time between payments. To calculate interest alone, simply subtract the principal from [latex]M[/latex].
  • exponential function : Any function in which an independent variable is in the form of an exponent; they are the inverse functions of logarithms.
  • compound interest : Interest, as on a loan or a bank account, that is calculated on the total on the principal plus accumulated unpaid interest.
  • Interest : The price paid for obtaining, or price received for providing, money or goods in a credit transaction, calculated as a fraction of the amount or value of what was borrowed.

Fundamentally, compound interest is an application of exponential functions that is found very commonly in every day life. Interest is, generally, a fee charged for the borrowing of money. The two classic cases are (1) interest accrued as part of loan and (2) interest accrued in a savings or other account. In the first case the client owes the amount borrowed plus the interest. In the second, the bank pays the client interest for maintaining money in the account. If you are the client, you are losing money in the first case and earning money in the second.

The amount of interest accrued depends on the principal (amount borrowed/deposited), the interest rate (a percentage of the principal), period (amount of time between interest payments) and time elapsed.

For questions of interest, unless otherwise stated, the amount borrowed/deposited remains unchanged. The client does not, for example, add or withdraw funds from the savings account after the initial deposit (the principal) is made.

Simple vs. Compound Interest

There exist two kinds of interest: simple and compound.

In simple interest, interest is accrued on the principal alone. This means that the amount of interest earned in each compounding period is the same because interest is earned based on the principal which remains unchanged.

In compound interest, interest is accrued on both the principal and on prior interest earned. For this reason, if all other conditions are the same (principal, rate, time elapsed and frequency of interest payments) compound interest grows at a faster rate than simple interest. The amount of interest earned increases with each compounding period.

Simple Interest: An Example

Simple interest is accrued linearly based on the formula:

[latex]\displaystyle I=p\cdot r \cdot \frac{t}{f}[/latex]

Where [latex]I[/latex] represents the interest, [latex]p[/latex] is the principal, [latex]r[/latex] is interest rate (expressed as a decimal), [latex]t[/latex] is time elapsed, and [latex]f[/latex] is the time elapsed per interest payment. The ratio of [latex]t[/latex] to [latex]f[/latex] is often simplified to the number of interest payments. Total amount owed/earned which includes the principal and the interest is given by [latex]A=P+I[/latex].

Example 1: You deposit $100 into a bank account earning 5% annual interest. How much money is in the account at the end of 1, 2, 3, 4, and [latex]n[/latex]years?

Let us begin by determining the interest at the end of the first year. We use the formula [latex]I=p\cdot r \cdot \frac{t}{f}[/latex] with [latex]p=100[/latex], [latex]r=.05[/latex], [latex]t=1[/latex] and [latex]f=1[/latex]. The interest earned at the end of the year is:

[latex]\displaystyle \begin{align} I&=(100)(.05)(\frac{1}{1})\\& =5 \end{align}[/latex]

The amount in the account at the conclusion of the year is given by [latex]100+5=105[/latex]. It is useful to note that the account will earn $[latex]5[/latex] in interest every single year irrespective of how long the money is in the account or what the amount in the account is during any given year. This can be seen by the fact that the amount after [latex]n[/latex] years is given by a linear function with slope equal to five (see table below).

The table below shows the calculations, interest earned and total amount in the account after 1, 2, 3, 4, and n years.

image

Calculations of interest earned and amount in the account for Example 1

Compound Interest: An Example

Compound interest is not linear, but exponential in form. That is, the amount of interest earned is not constant but instead changes with time based on the total amount of the amount in the account.

The equation representing investment value as a function of principal, interest rate, period and time is:

[latex]\displaystyle M=p(1+r)^\frac{t}{f}[/latex]

Where [latex]M[/latex] represents the total value (including principal), [latex]p[/latex] represents principal, [latex]r[/latex] is interest rate (expressed as a decimal), [latex]t[/latex] is time elapsed, and [latex]f[/latex] is the length of time between payments. To calculate interest alone, simply subtract the principal from [latex]M[/latex].

We now re-consider Example 1 above. This time we use compound interest instead.

Example 2: You deposit $100 into a bank account earning 5% interest compounded annually. How much money is in the account at the end of 1, 2, 3, 4, and [latex]n[/latex] years?

Let us begin by determining the amount in the account after the first year  using the formula [latex]M=p(1+r)^\frac{t}{f}[/latex] with [latex]p=100[/latex], [latex]r=.05[/latex], [latex]t=1[/latex] and [latex]f=1[/latex]. The amount in the account after one year is:

[latex]\displaystyle \begin{align} M&=(100)(1+.05)^\frac{1}{1}\\ &=105 \end{align}[/latex]

This is the exact amount that was in the account after the first year using simple interest. That is because the difference is that compound interest earns interest on both the principal and prior interest. At the end of the first year there was no prior interest in the account. We will see differences between simple and compound interest in this, and similar problems, in the second year.

Let us determine the amount in the account after the second year again using the formula [latex]M=p(1+r)^\frac{t}{f}[/latex] but now letting [latex]t=2[/latex]. We obtain:

[latex]\displaystyle \begin{align} M&=100(1+.05)^\frac{2}{1}\\ &=110.25 \end{align}[/latex]

That is, there are [latex]25[/latex] cents more in account in the second year using compound interest instead of simple interest. This might not seem like a lot but the amount of interest earned will continue to increase each year as there is more and more money in the account. Every year the interest earned will be higher than in the previous year, whereas in simple interest the amount each year is fixed.

The following table includes the calculations, interest earned and total amount in the account at the end of [latex]1,2,3,4[/latex] and [latex]n[/latex] years.

image

Calculations of interest earned and amount in the account for Example 2

The amount in the account is greater each year beginning with year two when using compound interest rather than simple interest. However, because the principal is so small and the number of years elapsed only 4, it does not appear that the difference between the two example is that great. Let us consider one last problem where we let the time elapsed be much greater.

Simple vs. Compound Interest Over Time: An Example

The biggest differences between the amount of money in an account using simple versus compound interest are seen over extended periods of time. To highlight this, we return the the examples we did prior and now consider how much money is in each account after 50 years.

Example 3: You deposit $100 into a bank account earning 5% interest compounded annually. How much money is in the account at the end of 50 years? How does this compare to the amount in the account after 50 years if the interest had been compounded annually?

The money in the account using simple interest after [latex]50[/latex] years is given by

[latex]\displaystyle \begin{align} P+I&=100+(100)(.05)(\frac{50}{1})\\ &=350 \end{align}[/latex]

However, the amount in the account using compound interest after [latex]50[/latex] years is given by:

[latex]\displaystyle \begin{align} M&=(100)(1+.05)^\frac{50}{1}\\ &=1146.74 \end{align}[/latex]

A difference of $[latex]796.74[/latex].

Frequency of Compounding Periods

The more frequent the compounding periods the more interest is accrued. Therefore, if one deposits money into a bank account that earns compound interest, and does not add or withdraw any additional funds, the amount of money in the bank would increase as the number of compounding periods per year increases. You earn more interest when interest is compounded quarterly ([latex]4[/latex] times a year) than once a year. You earn more interest when interest is compounded monthly ([latex]12[/latex] times a year) than quarterly. You earn more interest when interest is compounded daily ([latex]365[/latex] times a year) than monthly and so on. You earn the most interest when interest is compounded continuously.

Graph of interest accrued under differing number of compounding periods per year: Starting with a principal of $1000, interest rises exponentially. The graph shows that the more frequent the number of compounding periods the more interest is accrued and shows this visually for yearly, quarterly, monthly and continuous compounding.

Given that the more frequent the compounding periods per year, the more interest is accrued it might come as a surprise that money deposited into a bank account accrues compound interest continuously. You might expect the bank would choose a smaller number of compounding periods in order to pay out less in interest, but this is not the case. To see why let us take the following example: You deposit $[latex]1[/latex] in the bank in an account earning [latex]100[/latex]% interest. You do not add or withdraw money from the account. In this situation the amount of money in the account will be given by [latex](1+\frac{1}{n})^n[/latex] where [latex]n[/latex] is the number of compounding periods and [latex]\frac{1}{n}[/latex] is the rate per compounding period. This is a simplification of the prior formula used because of the specific conditions of this most recent situation. We expect that as [latex]n[/latex] increases the amount in the account also increases, but if the amount grows without bounds then banks would be giving away much money as they compound interest continuously.

The following table shows the amount in the account at the end of one year when interest is compounded with differing frequencies.

image

Amount in account after 1 year with interest compounded at different frequencies

What we see from the table is that while the interest earned increases as the number of compounding periods increase, that the rate at which it increases is slowing down. Going from [latex]1[/latex] to [latex]4[/latex] compounding periods yields an increase of [latex]44[/latex] cents but going from [latex]100[/latex] to [latex]1000[/latex] barely raises the interest [latex]1[/latex] cent. Another way of saying this is that the amount in the account is approaching a limiting value. The numbers in the table are getting closer and closer to the number [latex]e[/latex]. The number [latex]e[/latex] is used as the base of the natural logarithm and is equal to approximately [latex]2.718281828[/latex]. No matter how long the money is in the account, it will not grow beyond that value.

The formula for compound interest with the number of compounding periods going to infinity yields the formula for compounding continuously. It should come as no surprise that this formula involves the number [latex]e[/latex].

The formula for the amount of money in an account where interest is compounded continuously is given by [latex]A=Pe^{rt}[/latex] where [latex]P[/latex] is the principal, [latex]r[/latex] is the annual rate written as a decimal and [latex]t[/latex] is the time in years.

Exponential Decay

Exponential decay is the result of a function that decreases in proportion to its current value.

Use the exponential decay formula to calculate how much of something is left after a period of time

  • Exponential decrease can be modeled as: [latex]N(t)=N_0e^{-\lambda t}[/latex] where [latex]N[/latex] is the quantity, [latex]N_0[/latex] is the initial quantity, [latex]\lambda[/latex] is the decay constant, and [latex]t[/latex] is time.
  • Oftentimes, half-life is used to describe the amount of time required for half of a sample to decay. It can be defined mathematically as: [latex]t_{1/2}=\frac{ln(2)}{\lambda }[/latex]where [latex]t_{1/2}[/latex] is half-life.
  • Half-life can be inserted into the exponential decay model as such: [latex]N(t)=N_0(\frac{1}{2})^{t/t_{1/2}}[/latex]. Notice how the exponential changes, but the form of the function remains.
  • half-life : The time it takes for a substance (drug, radioactive nuclide, or other) to lose half of its pharmacological, physiological, biological, or radiological activity.
  • isotope : Any of two or more forms of an element where the atoms have the same number of protons, but a different number of neutrons. As a consequence, atoms for the same isotope will have the same atomic number but a different mass number (atomic weight).

Just as it is possible for a variable to grow exponentially as a function of another, so can the a variable decrease exponentially. Consider the decrease of a population that occurs at a rate proportional to its value. This rate at which the population is decreasing remains constant but as the population is continually decreasing the overall decline becomes less and less steep.

Exponential rate of change can be modeled algebraically by the following formula:

[latex]N(t)=N_0e^{-\lambda t}[/latex]

where [latex]N[/latex] is the quantity, [latex]N{_0}[/latex] is the initial quantity, [latex]\lambda[/latex] is the decay constant, and [latex]t[/latex] is time. The decay constant is indeed a constant, but the form of the equation (the negative exponent of e ) results in an ever-changing rate of decline.

The time it takes for a substance (drug, radioactive nuclide, or other) to lose half of its pharmacological, physiological, biological, or radiological activity is called its half-life. The exponential decay of the substance is a time-dependent decline and a prime example of exponential decay.

As an example let us assume we have a [latex]100[/latex] pounds of a substance with a half-life of [latex]5[/latex] years. Then in [latex]5[/latex] years half the amount ([latex]50[/latex] pounds) remains. In another [latex]5[/latex] years there will be [latex]25[/latex] pounds remaining. In another [latex]5[/latex] years, or [latex]15[/latex] years from the beginning, there will be [latex]12.5[/latex]. The amount by which the substance decreases, is itself slowly decreasing.

Use of Half Life in Carbon Dating

Half-life is very useful in determining the age of historical artifacts through a process known as carbon dating. Given a sample of carbon in an ancient, preserved piece of flesh, the age of the sample can be determined based on the percentage of radioactive carbon-13 remaining. 1.1% of carbon is C-13 and it decays to carbon-12.  C-13 has a half-life of 5700 years—that is, in 5700 years, half of a sample of C-13 will have converted to C-12, which represents approximately all the remaining carbon. Using this information it is possible to determine the age of the artifact given the amount of C-13 it presently contains, and comparing it to the amount of C-13 it should contain.

Half-life can be mathematically defined as:

[latex]\displaystyle t_{1/2}=\frac{ln(2)}{\lambda }[/latex]

It can also be conveniently inserted into the exponential decay formula as follows:

[latex]\displaystyle N(t)=N_0(\frac{1}{2})^{t/t_{1/2}}[/latex]

Thus, if a sample is found to contain 0.55% of its carbon as C-13 (exactly half of the usual 1.1%), it can be calculated that the sample has undergone exactly one half-life, and is thus 5,700 years old.

Below is a graph highlighting exponential decay of a radioactive substance. Using the graph, find that half-life.

image

Graph depicting radioactive decay: The amount of a substance undergoing radioactive decay decreases exponentially, eventually reaching zero. Since there is 50% of the substance left after 1 year, the half-life is 1 year.

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Mathematics LibreTexts

6.5: Applications of Exponential and Logarithmic Functions

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  • Page ID 80791

  • Carl Stitz & Jeff Zeager
  • Lakeland Community College & Lorain County Community College

As we mentioned in Section 6.1 , exponential and logarithmic functions are used to model a wide variety of behaviors in the real world. In the examples that follow, note that while the applications are drawn from many different disciplines, the mathematics remains essentially the same. Due to the applied nature of the problems we will examine in this section, the calculator is often used to express our answers as decimal approximations.

6.5.1. Applications of Exponential Functions

Perhaps the most well-known application of exponential functions comes from the financial world. Suppose you have \(\$ 100\) to invest at your local bank and they are offering a whopping \(5 \, \%\) annual percentage interest rate. This means that after one year, the bank will pay you \(5 \%\) of that \(\$100\), or \(\$ 100(0.05) =\$ 5\) in interest, so you now have \(\$105\). 1 This is in accordance with the formula for simple interest which you have undoubtedly run across at some point before.

Equation 6.1. Simple Interest

The amount of interest \(I\) accrued at an annual rate \(r\) on an investment a \(P\) after \(t\) years is \[I = Prt\nonumber\] The amount \(A\) in the account after \(t\) years is given by \[A = P + I = P + Prt = P(1+rt)\nonumber\]

Suppose, however, that six months into the year, you hear of a better deal at a rival bank. 2 Naturally, you withdraw your money and try to invest it at the higher rate there. Since six months is one half of a year, that initial \(\$100\) yields \(\$100(0.05)\left(\frac{1}{2}\right) = \$ 2.50\) in interest. You take your \(\$102.50\) off to the competitor and find out that those restrictions which may apply actually apply to you, and you return to your bank which happily accepts your \(\$102.50\) for the remaining six months of the year. To your surprise and delight, at the end of the year your statement reads \(\$105.06\), not \(\$105\) as you had expected. 3 Where did those extra six cents come from? For the first six months of the year, interest was earned on the original principal of \(\$100\), but for the second six months, interest was earned on \(\$102.50\), that is, you earned interest on your interest. This is the basic concept behind compound interest . In the previous discussion, we would say that the interest was compounded twice, or semiannually. 4 If more money can be earned by earning interest on interest already earned, a natural question to ask is what happens if the interest is compounded more often, say \(4\) times a year, which is every three months, or ‘quarterly.’ In this case, the money is in the account for three months, or \(\frac{1}{4}\) of a year, at a time. After the first quarter, we have \(A = P(1+rt) = \$100 \left(1 + 0.05 \cdot \frac{1}{4} \right) = \$101.25\). We now invest the \(\$101.25\) for the next three months and find that at the end of the second quarter, we have \(A = \$101.25 \left(1 + 0.05 \cdot \frac{1}{4} \right)\approx \$102.51\). Continuing in this manner, the balance at the end of the third quarter is \(\$103.79\), and, at last, we obtain \(\$105.08\). The extra two cents hardly seems worth it, but we see that we do in fact get more money the more often we compound. In order to develop a formula for this phenomenon, we need to do some abstract calculations. Suppose we wish to invest our principal \(P\) at an annual rate \(r\) and compound the interest \(n\) times per year. This means the money sits in the account \(\frac{1}{n}^{\mbox{\tiny th}}\) of a year between compoundings. Let \(A_{k}\) denote the amount in the account after the \(k^{\mbox{\tiny th}}\) compounding. Then \(A_{1} = P\left(1 + r\left(\frac{1}{n}\right)\right)\) which simplifies to \(A_{1} = P \left(1 + \frac{r}{n}\right)\). After the second compounding, we use \(A_{1}\) as our new principal and get \(A_{2} = A_{1} \left(1 + \frac{r}{n}\right) = \left[P \left(1 + \frac{r}{n}\right)\right]\left(1 + \frac{r}{n}\right) = P \left(1 + \frac{r}{n}\right)^2\). Continuing in this fashion, we get \(A_{3} =P \left(1 + \frac{r}{n}\right)^3\), \(A_{4} =P \left(1 + \frac{r}{n}\right)^4\), and so on, so that \(A_{k} = P \left(1 + \frac{r}{n}\right)^k\). Since we compound the interest \(n\) times per year, after \(t\) years, we have \(nt\) compoundings. We have just derived the general formula for compound interest below.

Equation 6.2. Compounded Interest

If an initial principal \(P\) is invested at an annual rate \(r\) and the interest is compounded \(n\) times per year, the amount \(A\) in the account after \(t\) years is \[A(t) = P \left(1 + \frac{r}{n}\right)^{nt}\nonumber\]

If we take \(P = 100\), \(r = 0.05\), and \(n = 4\), Equation 6.2 becomes \(A(t) = 100\left(1+ \frac{0.05}{4}\right)^{4t}\) which reduces to \(A(t) = 100(1.0125)^{4t}\). To check this new formula against our previous calculations, we find \(A\left(\frac{1}{4}\right) = 100(1.0125)^{4 \left(\frac{1}{4}\right)} = 101.25\), \(A\left(\frac{1}{2}\right) \approx \$102.51\), \(A\left(\frac{3}{4}\right) \approx \$103.79\), and \(A(1) \approx \$105.08\).

Example 6.5.1

Suppose \(\$2000\) is invested in an account which offers \(7.125 \%\) compounded monthly.

  • Express the amount \(A\) in the account as a function of the term of the investment \(t\) in years.
  • How much is in the account after \(5\) years?
  • How long will it take for the initial investment to double?
  • Find and interpret the average rate of change 5 of the amount in the account from the end of the fourth year to the end of the fifth year, and from the end of the thirty-fourth year to the end of the thirty-fifth year.
  • Substituting \(P = 2000\), \(r = 0.07125\), and \(n = 12\) (since interest is compounded monthly ) into Equation 6.2 yields \(A(t) = 2000\left(1 + \frac{0.07125}{12}\right)^{12t}=2000 (1.0059375)^{12t}\).
  • Since \(t\) represents the length of the investment in years, we substitute \(t=5\) into \(A(t)\) to find \(A(5) = 2000 (1.0059375)^{12(5)} \approx 2852.92\). After \(5\) years, we have approximately \(\$2852.92\).
  • Our initial investment is \(\$2000\), so to find the time it takes this to double, we need to find \(t\) when \(A(t) = 4000\). We get \(2000 (1.0059375)^{12t}=4000\), or \((1.0059375)^{12t}=2\). Taking natural logs as in Section 6.3 , we get \(t = \frac{\ln(2)}{12 \ln(1.0059375)} \approx 9.75\). Hence, it takes approximately \(9\) years \(9\) months for the investment to double.
  • To find the average rate of change of \(A\) from the end of the fourth year to the end of the fifth year, we compute \(\frac{A(5)-A(4)}{5-4} \approx 195.63\). Similarly, the average rate of change of \(A\) from the end of the thirty-fourth year to the end of the thirty-fifth year is \(\frac{A(35)-A(34)}{35-34} \approx 1648.21\). This means that the value of the investment is increasing at a rate of approximately \(\$195.63\) per year between the end of the fourth and fifth years, while that rate jumps to \(\$1648.21\) per year between the end of the thirty-fourth and thirty-fifth years. So, not only is it true that the longer you wait, the more money you have, but also the longer you wait, the faster the money increases. 6

We have observed that the more times you compound the interest per year, the more money you will earn in a year. Let’s push this notion to the limit. 7 Consider an investment of \(\$ 1\) invested at \(100 \%\) interest for \(1\) year compounded \(n\) times a year. Equation 6.2 tells us that the amount of money in the account after \(1\) year is \(A = \left(1+\frac{1}{n}\right)^{n}\). Below is a table of values relating \(n\) and \(A\).

\[\begin{array}{|r||r|} \hline n & A \\ \hline 1 & 2 \\ \hline 2 & 2.25 \\ \hline 4 & \approx 2.4414 \\ \hline 12 & \approx 2.6130 \\ \hline 360 & \approx 2.7145 \\ \hline 1000 & \approx 2.7169 \\ \hline 10000 & \approx 2.7181 \\ \hline 100000 & \approx 2.7182 \\ \hline \end{array}\nonumber\]

As promised, the more compoundings per year, the more money there is in the account, but we also observe that the increase in money is greatly diminishing. We are witnessing a mathematical ‘tug of war’. While we are compounding more times per year, and hence getting interest on our interest more often, the amount of time between compoundings is getting smaller and smaller, so there is less time to build up additional interest. With Calculus, we can show 8 that as \(n \rightarrow \infty\), \(A = \left(1+\frac{1}{n}\right)^{n} \rightarrow e\), where \(e\) is the natural base first presented in Section 6.1 . Taking the number of compoundings per year to infinity results in what is called continuously compounded interest.

Theorem 6.8.

If you invest \(\$1\) at \(100 \%\) interest compounded continuously, then you will have \(\$ e\) at the end of one year.

Using this definition of \(e\) and a little Calculus, we can take Equation 6.2 and produce a formula for continuously compounded interest.

Equation 6.3. Continuously Compounded Interest

If an initial principal \(P\) is invested at an annual rate \(r\) and the interest is compounded continuously, the amount \(A\) in the account after \(t\) years is \[A(t) = P e^{rt}\nonumber\]

If we take the scenario of Example 6.5.1 and compare monthly compounding to continuous compounding over \(35\) years, we find that monthly compounding yields \(A(35) = 2000 (1.0059375)^{12(35)}\) which is about \(\$ 24,\!035.28\), whereas continuously compounding gives \(A(35) = 2000e^{0.07125 (35)}\) which is about \(\$ 24,\!213.18\) - a difference of less than \(1 \%\).

Equations 6.2 and 6.3 both use exponential functions to describe the growth of an investment. Curiously enough, the same principles which govern compound interest are also used to model short term growth of populations. In Biology, The Law of Uninhibited Growth states as its premise that the instantaneous rate at which a population increases at any time is directly proportional to the population at that time. 9 In other words, the more organisms there are at a given moment, the faster they reproduce. Formulating the law as stated results in a differential equation, which requires Calculus to solve. Its solution is stated below.

Equation 6.4. Uninhibited Growth

If a population increases according to The Law of Uninhibited Growth, the number of organisms \(N\) at time \(t\) is given by the formula \[N(t) = N_0e^{kt},\nonumber\] where \(N(0) = N_0\) (read ‘\(N\) nought’) is the initial number of organisms and \(k>0\) is the constant of proportionality which satisfies the equation

\[\left(\mbox{instantaneous rate of change of $N(t)$ at time $t$}\right) = k \, N(t)\nonumber\]

It is worth taking some time to compare Equations 6.3 and 6.4 . In Equation 6.3 , we use \(P\) to denote the initial investment; in Equation 6.4 , we use \(N_0\) to denote the initial population. In Equation 6.3 , \(r\) denotes the annual interest rate, and so it shouldn’t be too surprising that the \(k\) in Equation 6.4 corresponds to a growth rate as well. While Equations 6.3 and 6.4 look entirely different, they both represent the same mathematical concept.

Example 6.5.2

In order to perform arthrosclerosis research, epithelial cells are harvested from discarded umbilical tissue and grown in the laboratory. A technician observes that a culture of twelve thousand cells grows to five million cells in one week. Assuming that the cells follow The Law of Uninhibited Growth, find a formula for the number of cells, \(N\), in thousands, after \(t\) days.

We begin with \(N(t) = N_0e^{kt}\). Since \(N\) is to give the number of cells in thousands , we have \(N_0 = 12\), so \(N(t) = 12e^{kt}\). In order to complete the formula, we need to determine the growth rate \(k\). We know that after one week, the number of cells has grown to five million. Since \(t\) measures days and the units of \(N\) are in thousands, this translates mathematically to \(N(7) = 5000\). We get the equation \(12e^{7k} = 5000\) which gives \(k = \frac{1}{7} \ln\left(\frac{1250}{3}\right)\). Hence, \(N(t) = 12e^{ \frac{t}{7} \ln\left(\frac{1250}{3}\right)}\). Of course, in practice, we would approximate \(k\) to some desired accuracy, say \(k \approx 0.8618\), which we can interpret as an \(86.18 \%\) daily growth rate for the cells.

Whereas Equations 6.3 and 6.4 model the growth of quantities, we can use equations like them to describe the decline of quantities. One example we’ve seen already is Example 6.1.1 in Section 6.1 . There, the value of a car declined from its purchase price of \(\$25,\!000\) to nothing at all. Another real world phenomenon which follows suit is radioactive decay. There are elements which are unstable and emit energy spontaneously. In doing so, the amount of the element itself diminishes. The assumption behind this model is that the rate of decay of an element at a particular time is directly proportional to the amount of the element present at that time. In other words, the more of the element there is, the faster the element decays. This is precisely the same kind of hypothesis which drives The Law of Uninhibited Growth, and as such, the equation governing radioactive decay is hauntingly similar to Equation 6.4 with the exception that the rate constant \(k\) is negative.

Equation 6.5. Radioactive Decay

The amount of a radioactive element \(A\) at time \(t\) is given by the formula \[A(t) = A_0e^{kt},\nonumber\] where \(A(0) = A_0\) is the initial amount of the element and \(k<0\) is the constant of proportionality which satisfies the equation

\[\left(\mbox{instantaneous rate of change of $A(t)$ at time $t$}\right) = k \, A(t)\nonumber\]

Example 6.5.3

Iodine-131 is a commonly used radioactive isotope used to help detect how well the thyroid is functioning. Suppose the decay of Iodine-131 follows the model given in Equation 6.5 , and that the half-life 10 of Iodine-131 is approximately \(8\) days. If \(5\) grams of Iodine-131 is present initially, find a function which gives the amount of Iodine-131, \(A\), in grams, \(t\) days later.

Since we start with \(5\) grams initially, Equation 6.5 gives \(A(t) = 5e^{kt}\). Since the half-life is \(8\) days, it takes \(8\) days for half of the Iodine-131 to decay, leaving half of it behind. Hence, \(A(8) = 2.5\) which means \(5e^{8k} = 2.5\). Solving, we get \(k = \frac{1}{8} \ln\left(\frac{1}{2}\right) = -\frac{\ln(2)}{8} \approx -0.08664\), which we can interpret as a loss of material at a rate of \(8.664 \%\) daily. Hence, \(A(t) = 5 e^{-\frac{t\ln(2)}{8}} \approx 5 e^{-0.08664t}\).

We now turn our attention to some more mathematically sophisticated models. One such model is Newton’s Law of Cooling, which we first encountered in Example 6.1.2 of Section 6.1 . In that example we had a cup of coffee cooling from \(160^{\circ}\mbox{F}\) to room temperature \(70^{\circ}\mbox{F}\) according to the formula \(T(t) = 70 + 90 e^{-0.1 t}\), where \(t\) was measured in minutes. In this situation, we know the physical limit of the temperature of the coffee is room temperature, 11 and the differential equation which gives rise to our formula for \(T(t)\) takes this into account. Whereas the radioactive decay model had a rate of decay at time \(t\) directly proportional to the amount of the element which remained at time \(t\), Newton’s Law of Cooling states that the rate of cooling of the coffee at a given time \(t\) is directly proportional to how much of a temperature exists between the coffee at time \(t\) and room temperature, not the temperature of the coffee itself. In other words, the coffee cools faster when it is first served, and as its temperature nears room temperature, the coffee cools ever more slowly. Of course, if we take an item from the refrigerator and let it sit out in the kitchen, the object’s temperature will rise to room temperature, and since the physics behind warming and cooling is the same, we combine both cases in the equation below.

Equation 6.6. Newton’s Law of Cooling (Warming)

The temperature \(T\) of an object at time \(t\) is given by the formula \[T(t) = T_{a} + \left(T_0 - T_{a}\right) e^{-kt},\nonumber\] where \(T(0) = T_0\) is the initial temperature of the object, \(T_{a}\) is the ambient temperature a and \(k>0\) is the constant of proportionality which satisfies the equation

\[\left(\mbox{instantaneous rate of change of $T(t)$ at time $t$}\right) = k \, \left(T(t) - T_{a}\right)\nonumber\]

If we re-examine the situation in Example 6.1.2 with \(T_0 = 160\), \(T_{a} = 70\), and \(k = 0.1\), we get, according to Equation 6.6 , \(T(t) = 70 + (160 - 70)e^{-0.1t}\) which reduces to the original formula given. The rate constant \(k = 0.1\) indicates the coffee is cooling at a rate equal to \(10 \%\) of the difference between the temperature of the coffee and its surroundings. Note in Equation 6.6 that the constant \(k\) is positive for both the cooling and warming scenarios. What determines if the function \(T(t)\) is increasing or decreasing is if \(T_0\) (the initial temperature of the object) is greater than \(T_{a}\) (the ambient temperature) or vice-versa, as we see in our next example.

Example 6.5.4.

A \(40^{\circ}\mbox{F}\) roast is cooked in a \(350^{\circ}\mbox{F}\) oven. After \(2\) hours, the temperature of the roast is \(125^{\circ}\mbox{F}\).

  • Assuming the temperature of the roast follows Newton’s Law of Warming, find a formula for the temperature of the roast \(T\) as a function of its time in the oven, \(t\), in hours.
  • The roast is done when the internal temperature reaches \(165^{\circ}\mbox{F}\). When will the roast be done?
  • The initial temperature of the roast is \(40^{\circ}\mbox{F}\), so \(T_0 = 40\). The environment in which we are placing the roast is the \(350^{\circ}\mbox{F}\) oven, so \(T_{a} = 350\). Newton’s Law of Warming tells us \(T(t) = 350 + (40-350)e^{-kt}\), or \(T(t) = 350 - 310e^{-kt}\). To determine \(k\), we use the fact that after \(2\) hours, the roast is \(125^{\circ}\mbox{F}\), which means \(T(2) = 125\). This gives rise to the equation \(350 - 310e^{-2k} = 125\) which yields \(k = -\frac{1}{2} \ln \left( \frac{45}{62} \right) \approx 0.1602\). The temperature function is \[T(t) = 350 - 310 e^{\frac{t}{2} \ln \left( \frac{45}{62} \right)} \approx 350- 310 e^{-0.1602 t}.\nonumber\]
  • To determine when the roast is done, we set \(T(t) = 165\). This gives \(350- 310 e^{-0.1602 t} = 165\) whose solution is \(t = -\frac{1}{0.1602} \ln \left( \frac{37}{62} \right) \approx 3.22\). It takes roughly \(3\) hours and \(15\) minutes to cook the roast completely.

If we had taken the time to graph \(y=T(t)\) in Example 6.5.4 , we would have found the horizontal asymptote to be \(y = 350\), which corresponds to the temperature of the oven. We can also arrive at this conclusion by applying a bit of ‘number sense’. As t \(\rightarrow \infty,-0.1602 t \approx \text { very big }(-)\) so that \(e^{-0.1602 t} \approx \text { very small }(+)\). The larger the value of \(t\), the smaller \(e^{-0.1602 t}\) becomes so that \(T(t) \approx 350-\text { very small }(+)\), which indicates the graph of \(y=T(t)\) is approaching its horizontal asymptote \(y=350\) from below. Physically, this means the roast will eventually warm up to \(350^{\circ}\mbox{F}\). 12 The function \(T\) is sometimes called a limited growth model, since the function \(T\) remains bounded as \(t \rightarrow \infty\). If we apply the principles behind Newton’s Law of Cooling to a biological example, it says the growth rate of a population is directly proportional to how much room the population has to grow. In other words, the more room for expansion, the faster the growth rate. The logistic growth model combines The Law of Uninhibited Growth with limited growth and states that the rate of growth of a population varies jointly with the population itself as well as the room the population has to grow.

Equation 6.7. Logistic Growth

If a population behaves according to the assumptions of logistic growth, the number of organisms \(N\) at time \(t\) is given by the equation \[N(t) =\dfrac{L}{1 + Ce^{-kLt}},\nonumber\] where \(N(0) = N_0\) is the initial population, \(L\) is the limiting population, a \(C\) is a measure of how much room there is to grow given by \[C = \dfrac{L}{N_0} - 1.\nonumber\] and \(k > 0\) is the constant of proportionality which satisfies the equation

\[\left(\mbox{instantaneous rate of change of $N(t)$ at time $t$}\right) = k \, N(t) \left(L - N(t)\right)\nonumber\]

The logistic function is used not only to model the growth of organisms, but is also often used to model the spread of disease and rumors. 13

Example 6.5.5

The number of people \(N\), in hundreds, at a local community college who have heard the rumor ‘Carl is afraid of Virginia Woolf’ can be modeled using the logistic equation

\[N(t) = \dfrac{84}{1+2799e^{-t}},\nonumber\]

where \(t\geq 0\) is the number of days after April 1, 2009.

  • Find and interpret \(N(0)\).
  • Find and interpret the end behavior of \(N(t)\).
  • How long until \(4200\) people have heard the rumor?
  • Check your answers to 2 and 3 using your calculator.
  • We find \(N(0) = \frac{84}{1+2799e^{0}} = \frac{84}{2800} = \frac{3}{100}\). Since \(N(t)\) measures the number of people who have heard the rumor in hundreds, \(N(0)\) corresponds to \(3\) people. Since \(t=0\) corresponds to April 1, 2009, we may conclude that on that day, \(3\) people have heard the rumor. 14
  • We could simply note that \(N(t)\) is written in the form of Equation 6.7 , and identify \(L = 84\). However, to see why the answer is \(84\), we proceed analytically. Since the domain of \(N\) is restricted to \(t \geq 0\), the only end behavior of significance is \(t \rightarrow \infty\). As we’ve seen before, 15 as \(t \rightarrow \infty\), we have \(1997 e^{-t} \rightarrow 0^{+}\) and so \(N(t) \approx \frac{84}{1+\text { very small }(+)} \approx 84\). Hence, as \(t \rightarrow \infty\), \(N(t) \rightarrow 84\). This means that as time goes by, the number of people who will have heard the rumor approaches \(8400\).
  • To find how long it takes until \(4200\) people have heard the rumor, we set \(N(t) = 42\). Solving \(\frac{84}{1+2799e^{-t}} = 42\) gives \(t = \ln(2799) \approx 7.937\). It takes around \(8\) days until \(4200\) people have heard the rumor.

Screen Shot 2022-04-18 at 7.12.22 PM.png

If we take the time to analyze the graph of \(y=N(x)\) above, we can see graphically how logistic growth combines features of uninhibited and limited growth. The curve seems to rise steeply, then at some point, begins to level off. The point at which this happens is called an inflection point or is sometimes called the ‘point of diminishing returns’. At this point, even though the function is still increasing, the rate at which it does so begins to decline. It turns out the point of diminishing returns always occurs at half the limiting population. (In our case, when \(y=42\).) While these concepts are more precisely quantified using Calculus, below are two views of the graph of \(y=N(x)\), one on the interval \([0,8]\), the other on \([8,15]\). The former looks strikingly like uninhibited growth; the latter like limited growth.

Screen Shot 2022-04-18 at 7.19.06 PM.png

6.5.2 Applications of Logarithms

Just as many physical phenomena can be modeled by exponential functions, the same is true of logarithmic functions. In Exercises 75 , 76 and 77 of Section 6.1 , we showed that logarithms are useful in measuring the intensities of earthquakes (the Richter scale), sound (decibels) and acids and bases (pH). We now present yet a different use of the a basic logarithm function, password strength .

Example 6.5.6

The information entropy \(H\), in bits, of a randomly generated password consisting of \(L\) characters is given by \(H = L \log_{2}(N)\), where \(N\) is the number of possible symbols for each character in the password. In general, the higher the entropy, the stronger the password.

  • If a \(7\) character case-sensitive 16 password is comprised of letters and numbers only, find the associated information entropy.
  • How many possible symbol options per character is required to produce a \(7\) character password with an information entropy of \(50\) bits?
  • There are \(26\) letters in the alphabet, \(52\) if upper and lower case letters are counted as different. There are \(10\) digits (\(0\) through \(9\)) for a total of \(N=62\) symbols. Since the password is to be \(7\) characters long, \(L = 7\). Thus, \(H = 7 \log_{2}(62) = \frac{7 \ln(62)}{\ln(2)} \approx 41.68\).
  • We have \(L = 7\) and \(H=50\) and we need to find \(N\). Solving the equation \(50 = 7 \log_{2}(N)\) gives \(N = 2^{50/7} \approx 141.323\), so we would need \(142\) different symbols to choose from. 17

Chemical systems known as buffer solutions have the ability to adjust to small changes in acidity to maintain a range of pH values. Buffer solutions have a wide variety of applications from maintaining a healthy fish tank to regulating the pH levels in blood. Our next example shows how the pH in a buffer solution is a little more complicated than the pH we first encountered in Exercise 77 in Section 6.1 .

Example 6.5.7

Blood is a buffer solution. When carbon dioxide is absorbed into the bloodstream it produces carbonic acid and lowers the pH. The body compensates by producing bicarbonate, a weak base to partially neutralize the acid. The equation 18 which models blood pH in this situation is \(\mbox{pH} = 6.1 + \log\left(\frac{800}{x} \right)\), where \(x\) is the partial pressure of carbon dioxide in arterial blood, measured in torr. Find the partial pressure of carbon dioxide in arterial blood if the pH is \(7.4\).

We set \(\mbox{pH} = 7.4\) and get \(7.4 = 6.1 + \log\left(\frac{800}{x} \right)\), or \(\log\left(\frac{800}{x} \right) = 1.3\). Solving, we find \(x = \frac{800}{10^{1.3}} \approx 40.09\). Hence, the partial pressure of carbon dioxide in the blood is about \(40\) torr.

Another place logarithms are used is in data analysis. Suppose, for instance, we wish to model the spread of influenza A (H1N1), the so-called ‘Swine Flu’. Below is data taken from the World Health Organization (WHO) where \(t\) represents the number of days since April 28, 2009, and \(N\) represents the number of confirmed cases of H1N1 virus worldwide.

\[\begin{array}{|c||c|c|c|c|c|c|c|c|c|c|c|c|c|} \hline t & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 \\ \hline N & 148 & 257 & 367 & 658 & 898 & 1085 & 1490 & 1893 & 2371 & 2500 & 3440 & 4379 & 4694 \\ \hline \end{array}\nonumber\]

\[\begin{array}{|c||c||c|c|c|c|c|c|} \hline t & 14 & 15 & 16 & 17 & 18 & 19& 20 \\ \hline N & 5251 & 5728 & 6497 & 7520 & 8451 & 8480 & 8829 \\ \hline \end{array}\nonumber\]

Making a scatter plot of the data treating \(t\) as the independent variable and \(N\) as the dependent variable gives

Screen Shot 2022-04-18 at 7.46.14 PM.png

Which models are suggested by the shape of the data? Thinking back Section 2.5 , we try a Quadratic Regression, with pretty good results.

Screen Shot 2022-04-18 at 7.49.10 PM.png

However, is there any scientific reason for the data to be quadratic? Are there other models which fit the data equally well, or better? Scientists often use logarithms in an attempt to ‘linearize’ data sets - in other words, transform the data sets to produce ones which result in straight lines. To see how this could work, suppose we guessed the relationship between \(N\) and \(t\) was some kind of power function, not necessarily quadratic, say \(N = B t^{A}\). To try to determine the \(A\) and \(B\), we can take the natural log of both sides and get \(\ln(N) = \ln\left(B t^{A}\right)\). Using properties of logs to expand the right hand side of this equation, we get \(\ln(N) = A \ln(t) + \ln(B)\). If we set \(X = \ln(t)\) and \(Y = \ln(N)\), this equation becomes \(Y = AX + \ln(B)\). In other words, we have a line with slope \(A\) and \(Y\)-intercept \(\ln(B)\). So, instead of plotting \(N\) versus \(t\), we plot \(\ln(N)\) versus \(\ln(t)\).

\[\begin{array}{|c||c|c|c|c|c|c|c|c|c|c|c|c|c|} \hline \ln(t) & 0 & 0.693 & 1.099 & 1.386& 1.609 & 1.792 & 1.946 & 2.079 & 2.197 & 2.302 & 2.398 & 2.485 & 2.565 \\ \hline \ln(N) & 4.997 & 5.549 & 5.905 & 6.489 & 6.800 & 6.989 & 7.306 & 7.546 & 7.771 & 7.824 & 8.143 & 8.385 & 8.454 \\ \hline \end{array}\nonumber\]

\[\begin{array}{|c||c||c|c|c|c|c|c|} \hline \ln(t) & 2.639 & 2.708 & 2.773 & 2.833 & 2.890 & 2.944 & 2.996 \\ \hline \ln(N) & 8.566 & 8.653 & 8.779 & 8.925 & 9.042 & 9.045 & 9.086 \\ \hline \end{array}\nonumber\]

Running a linear regression on the data gives

Screen Shot 2022-04-18 at 7.50.26 PM.png

The slope of the regression line is \(a \approx 1.512\) which corresponds to our exponent \(A\). The \(y\)-intercept \(b \approx 4.513\) corresponds to \(\ln(B)\), so that \(B \approx 91.201\). Hence, we get the model \(N = 91.201 t^{1.512}\), something from Section 5.3 . Of course, the calculator has a built-in ‘Power Regression’ feature. If we apply this to our original data set, we get the same model we arrived at before. 19

Screen Shot 2022-04-18 at 7.52.22 PM.png

This is all well and good, but the quadratic model appears to fit the data better, and we’ve yet to mention any scientific principle which would lead us to believe the actual spread of the flu follows any kind of power function at all. If we are to attack this data from a scientific perspective, it does seem to make sense that, at least in the early stages of the outbreak, the more people who have the flu, the faster it will spread, which leads us to proposing an uninhibited growth model. If we assume \(N = B e^{At}\) then, taking logs as before, we get \(\ln(N) = At + \ln(B)\). If we set \(X = t\) and \(Y = \ln(N)\), then, once again, we get \(Y = AX + \ln(B)\), a line with slope \(A\) and \(Y\)-intercept \(\ln(B)\). Plotting \(\ln(N)\) versus \(t\) gives the following linear regression.

Screen Shot 2022-04-18 at 7.53.05 PM.png

We see the slope is \(a \approx 0.202\) and which corresponds to \(A\) in our model, and the \(y\)-intercept is \(b \approx 5.596\) which corresponds to \(\ln(B)\). We get \(B \approx 269.414\), so that our model is \(N = 269.414e^{0.202t}\). Of course, the calculator has a built-in ‘Exponential Regression’ feature which produces what appears to be a different model \(N = 269.414 (1.22333419)^{t}\). Using properties of exponents, we write \(e^{0.202t} = \left(e^{0.202}\right)^t \approx (1.223848)^{t}\), which, had we carried more decimal places, would have matched the base of the calculator model exactly.

Screen Shot 2022-04-18 at 7.55.48 PM.png

The exponential model didn’t fit the data as well as the quadratic or power function model, but it stands to reason that, perhaps, the spread of the flu is not unlike that of the spread of a rumor and that a logistic model can be used to model the data. The calculator does have a ‘Logistic Regression’ feature, and using it produces the model \(N = \frac{10739.147}{1 + 42.416 e^{0.268 t}}\).

Screen Shot 2022-04-18 at 7.56.22 PM.png

This appears to be an excellent fit, but there is no friendly coefficient of determination, \(R^2\), by which to judge this numerically. There are good reasons for this, but they are far beyond the scope of the text. Which of the models, quadratic, power, exponential, or logistic is the ‘best model’? If by ‘best’ we mean ‘fits closest to the data,’ then the quadratic and logistic models are arguably the winners with the power function model a close second. However, if we think about the science behind the spread of the flu, the logistic model gets an edge. For one thing, it takes into account that only a finite number of people will ever get the flu (according to our model, \(10,\!739\)), whereas the quadratic model predicts no limit to the number of cases. As we have stated several times before in the text, mathematical models, regardless of their sophistication, are just that: models, and they all have their limitations. 20

6.5.3. Exercises

For each of the scenarios given in Exercises 1 - 6,

  • Find the amount \(A\) in the account as a function of the term of the investment \(t\) in years.
  • Determine how much is in the account after \(5\) years, \(10\) years, \(30\) years and \(35\) years. Round your answers to the nearest cent.
  • Determine how long will it take for the initial investment to double. Round your answer to the nearest year.
  • Find and interpret the average rate of change of the amount in the account from the end of the fourth year to the end of the fifth year, and from the end of the thirty-fourth year to the end of the thirty-fifth year. Round your answer to two decimal places.
  • \(\$500\) is invested in an account which offers \(0.75 \%\), compounded monthly.
  • \(\$500\) is invested in an account which offers \(0.75 \%\), compounded continuously.
  • \(\$1000\) is invested in an account which offers \(1.25 \%\), compounded monthly.
  • \(\$1000\) is invested in an account which offers \(1.25 \%\), compounded continuously.
  • \(\$5000\) is invested in an account which offers \(2.125 \%\), compounded monthly.
  • \(\$5000\) is invested in an account which offers \(2.125 \%\), compounded continuously.
  • Look back at your answers to Exercises 1 - 6. What can be said about the difference between monthly compounding and continuously compounding the interest in those situations? With the help of your classmates, discuss scenarios where the difference between monthly and continuously compounded interest would be more dramatic. Try varying the interest rate, the term of the investment and the principal. Use computations to support your answer.
  • How much money needs to be invested now to obtain \(\$2000\) in 3 years if the interest rate in a savings account is \(0.25 \%\), compounded continuously? Round your answer to the nearest cent.
  • How much money needs to be invested now to obtain \(\$5000\) in 10 years if the interest rate in a CD is \(2.25 \%\), compounded monthly? Round your answer to the nearest cent.
  • If \(P = 2000\) what is \(A(8)\)?
  • Solve the equation \(A(t) = 4000\) for \(t\).
  • What principal \(P\) should be invested so that the account balance is $2000 is three years?
  • What principal \(P\) should be invested so that the account balance is $2000 in three years?
  • The Annual Percentage Yield is the interest rate that returns the same amount of interest after one year as the compound interest does. With the help of your classmates, compute the APY for this investment.
  • A finance company offers a promotion on \(\$5000\) loans. The borrower does not have to make any payments for the first three years, however interest will continue to be charged to the loan at \(29.9 \%\) compounded continuously. What amount will be due at the end of the three year period, assuming no payments are made? If the promotion is extended an additional three years, and no payments are made, what amount would be due?
  • Use Equation 6.2 to show that the time it takes for an investment to double in value does depend on the principal \(P\), but rather, depends only on the APR and the number of compoundings per year. Let \(n = 12\) and with the help of your classmates compute the doubling time for a variety of rates \(r\). Then look up the Rule of 72 and compare your answers to what that rule says. If you’re really interested 21 in Financial Mathematics, you could also compare and contrast the Rule of 72 with the Rule of 70 and the Rule of 69.

In Exercises 14 - 18, we list some radioactive isotopes and their associated half-lives. Assume that each decays according to the formula \(A(t)=A_{0} e^{k t}\) where \(A_{0}\) is the initial amount of the material and \(k\) is the decay constant. For each isotope:

  • Find the decay constant \(k\). Round your answer to four decimal places.
  • Find a function which gives the amount of isotope \(A\) which remains after time \(t\). (Keep the units of \(A\) and \(t\) the same as the given data.)
  • Determine how long it takes for \(90 \%\) of the material to decay. Round your answer to two decimal places. (HINT: If \(90 \%\) of the material decays, how much is left?)
  • Cobalt 60, used in food irradiation, initial amount 50 grams, half-life of \(5.27\) years.
  • Phosphorus 32, used in agriculture, initial amount 2 milligrams, half-life \(14\) days.
  • Chromium 51, used to track red blood cells, initial amount 75 milligrams, half-life \(27.7\) days.
  • Americium 241, used in smoke detectors, initial amount 0.29 micrograms, half-life \(432.7\) years.
  • Uranium 235, used for nuclear power, initial amount \(1\) kg grams, half-life \(704\) million years.
  • With the help of your classmates, show that the time it takes for \(90 \%\) of each isotope listed in Exercises 14 - 18 to decay does not depend on the initial amount of the substance, but rather, on only the decay constant \(k\). Find a formula, in terms of \(k\) only, to determine how long it takes for \(90 \%\) of a radioactive isotope to decay.
  • In Example 6.1.1 in Section 6.1 , the exponential function \(V(x) = 25 \left(\frac{4}{5}\right)^{x}\) was used to model the value of a car over time. Use the properties of logs and/or exponents to rewrite the model in the form \(V(t) = 25e^{kt}\).
  • Find and interpret \(G(0)\).
  • According to the model, what should have been the GDP in 2007? In 2010? (According to the US Department of Commerce , the 2007 GDP was \(\$14,369.1\) billion and the 2010 GDP was \(\$14,657.8\) billion.)
  • What was the diameter of the tumor when it was originally detected?
  • How long until the diameter of the tumor doubles?
  • Find the growth constant \(k\). Round your answer to four decimal places.
  • Find a function which gives the number of bacteria \(N(t)\) after \(t\) minutes.
  • How long until there are 9000 bacteria? Round your answer to the nearest minute.
  • Find a function which gives the number of yeast (in millions) per cc \(N(t)\) after \(t\) hours.
  • What is the doubling time for this strain of yeast?
  • The Law of Uninhibited Growth also applies to situations where an animal is re-introduced into a suitable environment. Such a case is the reintroduction of wolves to Yellowstone National Park. According to the National Park Service , the wolf population in Yellowstone National Park was 52 in 1996 and 118 in 1999. Using these data, find a function of the form \(N(t)=N_{0} e^{k t}\) which models the number of wolves \(t\) years after 1996. (Use \(t = 0\) to represent the year 1996. Also, round your value of \(k\) to four decimal places.) According to the model, how many wolves were in Yellowstone in 2002? (The recorded number is 272.)
  • During the early years of a community, it is not uncommon for the population to grow according to the Law of Uninhibited Growth. According to the Painesville Wikipedia entry, in 1860, the Village of Painesville had a population of 2649. In 1920, the population was 7272. Use these two data points to fit a model of the form \(N(t)=N_{0} e^{k t}\) were \(N(t)\) is the number of Painesville Residents \(t\) years after 1860. (Use \(t = 0\) to represent the year 1860. Also, round the value of \(k\) to four decimal places.) According to this model, what was the population of Painesville in 2010? (The 2010 census gave the population as 19,563) What could be some causes for such a vast discrepancy? For more on this, see Exercise 37.
  • Find and interpret \(P(0)\).
  • Find the population of Sasquatch in Bigfoot county in 2013. Round your answer to the nearest Sasquatch.
  • When will the population of Sasquatch in Bigfoot county reach 60? Round your answer to the nearest year.
  • Find and interpret the end behavior of the graph of \(y = P(t)\). Check your answer using a graphing utility.
  • Use Equation 6.5 to express the amount of Carbon-14 left from an initial \(N\) milligrams as a function of time \(t\) in years.
  • What percentage of the original amount of Carbon-14 is left after 20,000 years?
  • If an old wooden tool is found in a cave and the amount of Carbon-14 present in it is estimated to be only 42% of the original amount, approximately how old is the tool?
  • Radiocarbon dating is not as easy as these exercises might lead you to believe. With the help of your classmates, research radiocarbon dating and discuss why our model is somewhat over-simplified.
  • Carbon-14 cannot be used to date inorganic material such as rocks, but there are many other methods of radiometric dating which estimate the age of rocks. One of them, Rubidium-Strontium dating, uses Rubidium-87 which decays to Strontium-87 with a half-life of 50 billion years. Use Equation 6.5 to express the amount of Rubidium-87 left from an initial 2.3 micrograms as a function of time \(t\) in billions of years. Research this and other radiometric techniques and discuss the margins of error for various methods with your classmates.
  • Use Equation 6.5 to show that \(k = -\dfrac{\ln(2)}{h}\) where \(h\) is the half-life of the radioactive isotope.
  • Express the temperature \(T\) (in \(^{\circ}\)F) as a function of time \(t\) (in minutes).
  • Find the time at which the roast would have dropped to \(140^{\circ}\)F had it not been carved and eaten.

\[y(x) = \frac{1}{4} x^2-\frac{1}{4} \ln(x)-\frac{1}{4}\nonumber\]

Use your calculator to graph this path for \(x > 0\). Describe the behavior of \(y\) as \(x \rightarrow 0^{+}\) and interpret this physically.

  • The current \(i\) measured in amps in a certain electronic circuit with a constant impressed voltage of 120 volts is given by \(i(t) = 2 - 2e^{-10t}\) where \(t \geq 0\) is the number of seconds after the circuit is switched on. Determine the value of \(i\) as \(t \rightarrow \infty\). (This is called the steady state current.)

\[i(t) = \left\{ \begin{array}{rcl} 2 - 2e^{-10t} & \mbox{if} & 0 \leq t < 30 \\ [6pt] \left(2 - 2e^{-300}\right) e^{-10t+300} & \mbox{if} & t \geq 30 \end{array} \right.\nonumber\]

With the help of your calculator, graph \(y = i(t)\) and discuss with your classmates the physical significance of the two parts of the graph \(0 \leq t < 30\) and \(t \geq 30\).

  • In Exercise 6a in Section 2.5 , we stated that the cable of a suspension bridge formed a parabola but that a free hanging cable did not. A free hanging cable forms a and its basic shape is given by \(y = \frac{1}{2}\left(e^{x} + e^{-x}\right)\). Use your calculator to graph this function. What are its domain and range? What is its end behavior? Is it invertible? How do you think it is related to the function given in Exercise 47 in Section 6.3 and the one given in the answer to Exercise 38 in Section 6.4 ? When flipped upside down, the catenary makes an arch. The Gateway Arch in St. Louis, Missouri has the shape \[y = 757.7 - \frac{127.7}{2}\left(e^{\frac{x}{127.7}} + e^{-\frac{x}{127.7}}\right)\nonumber\] where \(x\) and \(y\) are measured in feet and \(-315 \leq x \leq 315\). Find the highest point on the arch.

\(\begin{array}{|l|r|r|r|r|r|r|r|r|r|r|} \hline \text { Year } x & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & & 9 \\ \hline \text { Number of } & & & & & & & & & & \\ \text { Cats } N(x) & 12 & 66 & 382 & 2201 & 12680 & 73041 & 420715 & 2423316 & 13968290 & 80399780 \\ \hline \end{array}\)

  • Use a graphing utility to perform an exponential regression on the data from 1860 through 1920 only, letting \(t = 0\) represent the year 1860 as before. How does this calculator model compare with the model you found in Exercise 26? Use the calculator’s exponential model to predict the population in 2010. (The 2010 census gave the population as 19,563)
  • The logistic model fit to all of the given data points for the population of Painesville \(t\) years after 1860 (again, using \(t = 0\) as 1860) is \[P(t) = \dfrac{18691}{1+9.8505e^{-0.03617t}}\nonumber\] According to this model, what should the population of Painesville have been in 2010? (The 2010 census gave the population as 19,563.) What is the population limit of Painesville?
  • Use your calculator to fit a logistic model to these data, using \(x = 0\) to represent the year 1860.
  • Graph these data and your logistic function on your calculator to judge the reasonableness of the fit.
  • Use this model to estimate the population of Lake County in 2010. (The 2010 census gave the population to be 230,041.)
  • According to your model, what is the population limit of Lake County, Ohio?

With the help of your classmates, find a model for this data.

With the help of your classmates, find a model for this data. Unlike most of the phenomena we have studied in this section, there is no single differential equation which governs the enrollment growth. Thus there is no scientific reason to rely on a logistic function even though the data plot may lead us to that model. What are some factors which influence enrollment at a community college and how can you take those into account mathematically?

With the help of your classmates, find a model for this data and make a prediction for the Opening Day enrollment as well as the Day 15 enrollment. (WARNING: The registration period for 2009 was one week shorter than it was in 2008 so Opening Day would be \(x = 21\) and Day 15 is \(x = 23\).)

6.5.4. Answers

  • \(A(t) = 500\left(1 + \frac{0.0075}{12}\right)^{12t}\)
  • \(A(5) \approx \$ 519.10\), \(A(10) \approx \$ 538.93\), \(A(30) \approx \$ 626.12\), \(A(35) \approx \$ 650.03\)
  • It will take approximately \(92\) years for the investment to double.
  • The average rate of change from the end of the fourth year to the end of the fifth year is approximately \(3.88\). This means that the investment is growing at an average rate of \(\$3.88\) per year at this point. The average rate of change from the end of the thirty-fourth year to the end of the thirty-fifth year is approximately \(4.85\). This means that the investment is growing at an average rate of \(\$4.85\) per year at this point.
  • \(A(t) = 500e^{0.0075t}\)
  • \(A(5) \approx \$ 519.11\), \(A(10) \approx \$ 538.94\), \(A(30) \approx \$ 626.16\), \(A(35) \approx \$ 650.09\)
  • The average rate of change from the end of the fourth year to the end of the fifth year is approximately \(3.88\). This means that the investment is growing at an average rate of \(\$3.88\) per year at this point. The average rate of change from the end of the thirty-fourth year to the end of the thirty-fifth year is approximately \(4.86\). This means that the investment is growing at an average rate of \(\$4.86\) per year at this point.
  • \(A(t) = 1000\left(1 + \frac{0.0125}{12}\right)^{12t}\)
  • \(A(5) \approx \$ 1064.46\), \(A(10) \approx \$ 1133.07\), \(A(30) \approx \$ 1454.71\), \(A(35) \approx \$ 1548.48\)
  • It will take approximately \(55\) years for the investment to double.
  • The average rate of change from the end of the fourth year to the end of the fifth year is approximately \(13.22\). This means that the investment is growing at an average rate of \(\$13.22\) per year at this point. The average rate of change from the end of the thirty-fourth year to the end of the thirty-fifth year is approximately \(19.23\). This means that the investment is growing at an average rate of \(\$19.23\) per year at this point.
  • \(A(t) = 1000e^{0.0125t}\)
  • \(A(5) \approx \$ 1064.49\), \(A(10) \approx \$ 1133.15\), \(A(30) \approx \$ 1454.99\), \(A(35) \approx \$ 1548.83\)
  • The average rate of change from the end of the fourth year to the end of the fifth year is approximately \(13.22\). This means that the investment is growing at an average rate of \(\$13.22\) per year at this point. The average rate of change from the end of the thirty-fourth year to the end of the thirty-fifth year is approximately \(19.24\). This means that the investment is growing at an average rate of \(\$19.24\) per year at this point.
  • \(A(t) = 5000\left(1 + \frac{0.02125}{12}\right)^{12t}\)
  • \(A(5) \approx \$ 5559.98\), \(A(10) \approx \$ 6182.67\), \(A(30) \approx \$ 9453.40\), \(A(35) \approx \$ 10512.13\)
  • It will take approximately \(33\) years for the investment to double.
  • The average rate of change from the end of the fourth year to the end of the fifth year is approximately \(116.80\). This means that the investment is growing at an average rate of \(\$116.80\) per year at this point. The average rate of change from the end of the thirty-fourth year to the end of the thirty-fifth year is approximately \(220.83\). This means that the investment is growing at an average rate of \(\$220.83\) per year at this point.
  • \(A(t) = 5000e^{0.02125t}\)
  • \(A(5) \approx \$ 5560.50\), \(A(10) \approx \$ 6183.83\), \(A(30) \approx \$ 9458.73\), \(A(35) \approx \$ 10519.05\)
  • The average rate of change from the end of the fourth year to the end of the fifth year is approximately \(116.91\). This means that the investment is growing at an average rate of \(\$116.91\) per year at this point. The average rate of change from the end of the thirty-fourth year to the end of the thirty-fifth year is approximately \(221.17\). This means that the investment is growing at an average rate of \(\$221.17\) per year at this point.
  • \(P = \frac{2000}{e^{0.0025 \cdot 3}} \approx \$ 1985.06\)
  • \(P = \frac{5000}{\left(1 + \frac{0.0225}{12}\right)^{12 \cdot 10}} \approx \$ 3993.42\)
  • \(A(8) = 2000\left(1 + \frac{0.0025}{12}\right)^{12 \cdot 8} \approx \$2040.40\)
  • \(t = \dfrac{\ln(2)}{12 \ln\left(1 + \frac{0.0025}{12}\right)} \approx 277.29\) years
  • \(P = \dfrac{2000}{\left(1 + \frac{0.0025}{12}\right)^{36}} \approx \$1985.06\)
  • \(A(8) = 2000\left(1 + \frac{0.0225}{12}\right)^{12 \cdot 8} \approx \$2394.03\)
  • \(t = \dfrac{\ln(2)}{12 \ln\left(1 + \frac{0.0225}{12}\right)} \approx 30.83\) years
  • \(P = \dfrac{2000}{\left(1 + \frac{0.0225}{12}\right)^{36}} \approx \$1869.57\)
  • \(\left(1 + \frac{0.0225}{12}\right)^{12} \approx 1.0227\) so the APY is 2.27%
  • \(A(3) = 5000e^{0.299 \cdot 3} \approx \$12,226.18\), \(A(6) = 5000e^{0.299 \cdot 6} \approx \$30,067.29\)
  • \(k = \frac{\ln(1/2)}{5.27} \approx -0.1315\)
  • \(A(t) = 50e^{-0.1315t}\)
  • \(t = \frac{\ln(0.1)}{-0.1315} \approx 17.51\) years.
  • \(k = \frac{\ln(1/2)}{14} \approx -0.0495\)
  • \(A(t) = 2e^{-0.0495t}\)
  • \(t = \frac{\ln(0.1)}{-0.0495} \approx 46.52\) days.
  • \(k = \frac{\ln(1/2)}{27.7} \approx -0.0250\)
  • \(A(t) = 75e^{-0.0250t}\)
  • \(t = \frac{\ln(0.1)}{-0.025} \approx 92.10\) days.
  • \(k = \frac{\ln(1/2)}{432.7} \approx -0.0016\)
  • \(A(t) = 0.29e^{-0.0016t}\)
  • \(t = \frac{\ln(0.1)}{-0.0016} \approx 1439.11\) years.
  • \(k = \frac{\ln(1/2)}{704} \approx -0.0010\)
  • \(A(t) = e^{-0.0010t}\)
  • \(t = \frac{\ln(0.1)}{-0.0010} \approx 2302.58\) million years, or \(2.30\) billion years.
  • \(t = \frac{\ln(0.1)}{k} = -\frac{\ln(10)}{k}\)
  • \(V(t) = 25e^{\ln\left(\frac{4}{5}\right)t} \approx 25e^{-0.22314355t}\)
  • \(G(0) = 9743.77\) This means that the GDP of the US in 2000 was \(\$9743.77\) billion dollars.
  • \(G(7) = 13963.24\) and \(G(10) = 16291.25\), so the model predicted a GDP of \(\$ 13,963.24\) billion in 2007 and \(\$ 16,291.25\) billion in 2010.
  • \(D(0) = 15\), so the tumor was 15 millimeters in diameter when it was first detected.
  • \(t = \frac{\ln(2)}{0.0277} \approx 25\) days.
  • \(k = \frac{\ln(2)}{20} \approx 0.0346\)
  • \(N(t) = 1000e^{0.0346 t}\)
  • \(t = \frac{\ln(9)}{0.0346} \approx 63\) minutes
  • \(k = \frac{1}{2}\frac{\ln(6)}{2.5} \approx 0.4377\)
  • \(N(t) = 2.5e^{0.4377 t}\)
  • \(t = \frac{\ln(2)}{0.4377} \approx 1.58\) hours
  • \(N_{0}=52, k=\frac{1}{3} \ln \left(\frac{118}{52}\right) \approx 0.2731, N(t)=52 e^{0.2731 t} \cdot N(6) \approx 268\).
  • \(N_{0}=2649, k=\frac{1}{60} \ln \left(\frac{7272}{2649}\right) \approx 0.0168, N(t)=2649 e^{0.0168 t} \cdot N(150) \approx 32923\), so the population of Painesville in 2010 based on this model would have been 32,923.
  • \(P(0) = \frac{120}{4.167} \approx 29\). There are 29 Sasquatch in Bigfoot County in 2010.
  • \(P(3) = \frac{120}{1+3.167e^{-0.05(3)}} \approx 32\) Sasquatch.
  • \(t = 20 \ln(3.167) \approx 23\) years.
  • As \(t \rightarrow \infty\), \(P(t) \rightarrow 120\). As time goes by, the Sasquatch Population in Bigfoot County will approach 120. Graphically, \(y = P(x)\) has a horizontal asymptote \(y=120\).
  • \(A(t) = Ne^{-\left(\frac{\ln(2)}{5730}\right)t} \approx Ne^{-0.00012097t}\)
  • \(A(20000) \approx 0.088978 \cdot N\) so about 8.9% remains
  • \(t \approx \dfrac{\ln(.42)}{-0.00012097} \approx 7171\) years old
  • \(A(t) = 2.3e^{-0.0138629t}\)
  • \(T(t) = 75 + 105e^{-0.005005t}\)
  • The roast would have cooled to \(140^{\circ}\)F in about 95 minutes.

Screen Shot 2022-04-19 at 6.33.34 PM.png

  • The steady state current is 2 amps.

\(N(x) = 2.02869(5.74879)^{x} = 2.02869e^{1.74899x}\) with \(r^{2} \approx 0.999995\). This is also an excellent fit and corresponds to our linearized model because \(\ln(2.02869) \approx 0.70739\).

  • The calculator gives: \(y = 2895.06 (1.0147)^{x}\). Graphing this along with our answer from Exercise 26 over the interval \([0,60]\) shows that they are pretty close. From this model, \(y(150) \approx 25840\) which once again overshoots the actual data value.
  • \(P(150) \approx 18717\), so this model predicts 17,914 people in Painesville in 2010, a more conservative number than was recorded in the 2010 census. As \(t \rightarrow \infty\), \(P(t) \rightarrow 18691\). So the limiting population of Painesville based on this model is 18,691 people.
  • \(y = \dfrac{242526}{1+874.62e^{-0.07113x}}\), where \(x\) is the number of years since 1860.

Screen Shot 2022-04-21 at 4.08.08 PM.png

  • \(y(140) \approx 232889\), so this model predicts 232,889 people in Lake County in 2010.
  • As \(x \rightarrow \infty\), \(y \rightarrow 242526\), so the limiting population of Lake County based on this model is 242,526 people.

1 How generous of the

2 Some restrictions may apply.

3 Actually, the final balance should be $105.0625.

4 Using this convention, simple interest after one year is the same as compounding the interest only once.

5 See Definition 2.3 in section 2.1 .

6 In fact, the rate of increase of the amount in the account is exponential as well. This is the quality that really defines exponential functions and we refer the reader to a course in Calculus.

7 Once you’ve had a semester of Calculus, you’ll be able to fully appreciate this very lame pun.

8 Or define, depending on your point of view.

9 The average rate of change of a function over an interval was first introduced in Section 2.1 . Instantaneous rates of change are the business of Calculus, as is mentioned on Page 161.

10 The time it takes for half of the substance to decay.

11 The Second Law of Thermodynamics states that heat can spontaneously flow from a hotter object to a colder one, but not the other way around. Thus, the coffee could not continue to release heat into the air so as to cool below room temperature.

12 at which point it would be more toast than roast.

13 Which can be just as damaging as diseases.

14 Or, more likely, three people started the rumor. I’d wager Jeff, Jamie, and Jason started it. So much for telling your best friends something in confidence!

15 See, for example, Example 6.1.2 .

16 That is, upper and lower case letters are treated as different characters.

17 Since there are only 94 distinct ASCII keyboard characters, to achieve this strength, the number of characters in the password should be increase.

18 Derived from the Henderson-Hasselbalch Equation . See Exercise 43 in Section 6.2 . Hasselbalch himself was studying carbon dioxide dissolving in blood - a process called metabolic acidosis .

19 Critics may question why the authors of the book have chosen to even discuss linearization of data when the calculator has a Power Regression built-in and ready to go. Our response: talk to your science faculty

20 Speaking of limitations, as of June 3, 2009, there were 19,273 confirmed cases of influenza A (H1N1). This is well above our prediction of 10,739. Each time a new report is issued, the data set increases and the model must be recalculated. We leave this recalculation to the reader.

21 Awesome pun!

22 This roast was enjoyed by Jeff and his family on June 10, 2009. This is real data, folks!

23 The authors thank Dr. Wendy Marley and her staff for this data and Dr. Marcia Ballinger for the permission to use it in this problem.

IMAGES

  1. S23 Application of Exponential Functions (Example 2)

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  2. Examples of Applications of Exponential Functions

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  3. Lesson: Applications of Exponential Functions

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  4. Applications of Exponential Functions

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  5. 4.2 Applications of Exponential Functions

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  6. Exponential Functions Applications Notes and Practice

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VIDEO

  1. Sec 2 Algebra Exponential Function and its Applications Session 18

  2. AP Calculus 6.5: Accumulation Functions (Homework)

  3. Exponential functions exercise (SE)

  4. Tech Math 3: Section 4.1b

  5. Exponential Equation| How to solve exponential equations

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COMMENTS

  1. 15.1: Applications of exponential functions

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  2. Applications of Exponential Functions

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  4. 4.1e: Exercises

    10) The height of a projectile at time \(t\) is represented by the function \(h(t)= -4.9t^2 + 18t + 40\) \( \bigstar \) For the following exercises, determine whether the table could represent a function that is linear, exponential, or neither. If it appears to be exponential, find a function that passes through the points.

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  7. 6.1 Exponential Functions

    A function that models exponential growth grows by a rate proportional to the amount present. For any real number x and any positive real numbers a and b such that b ≠ 1, an exponential growth function has the form. f ( x) = a b x. where. a. a is the initial or starting value of the function. b.

  8. Applications of Exponential Functions

    Example. Suppose that a certain radioactive substance has a half-life of 20 years. If you initially have 100 g of this substance, then find the quantity function Q(t) of this substance after t years. Since Q(0) = 100 and h = 20, we have. Q(t) = 100( 1 2) t 20. I hope that this was helpful.

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    This topic covers: - Radicals & rational exponents - Graphs & end behavior of exponential functions - Manipulating exponential expressions using exponent properties - Exponential growth & decay - Modeling with exponential functions - Solving exponential equations - Logarithm properties - Solving logarithmic equations - Graphing logarithmic functions - Logarithmic scale

  10. Solved Identify 10 applications of exponential functions in

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    Graphing a Logistic Growth Model. Graphically, the logistic function resembles an exponential function followed by a logarithmic function that approaches a horizontal asymptote. This horizontal asymptote represents the carrying capacity. That is, \displaystyle y=c y = c is a horizontal asymptote of the graph.

  12. Algebra

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    Exponential and Logarithmic Equations and Applications . Steps for solving exponential equations: 1. Isolate the exponential expression on one side of the equation (if possible). 2. Take the log of both sides and "bring down the exponent" using the power property of logarithms. 3. Solve for the variable. RECALL: Properties of Logarithms

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    Question: Applications of Exponential Functions 10/10 answered Skill 1: Solve an exponential equation Q1 Solve an exponential equation Homework Answered Solve the equation. Round the answer to two decimal places. Enter the value oft below. 40 0.04 50.000 = 10.000 ( 1 + 4 Type your numeric answer and submit -2.74 Hint

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    Homework 10.2 Part two (Application of exponential functions) 3/12th -2021 Use the compounded continuously formula: A = pe" pert $900 invested for 5 years t 9% compounded continuously produces. $840 invested for 10 years at 15% compounded continuously produces. $3265 invested for 5 years t 9% compounded continuously produces. $3265 invested for 5 years t 9% compounded continuously produces.

  21. Solved Application of Exponential Function General

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  24. Homework Questions for Applications involving Exponential Functions

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