Contents

Preface
1Foundations
- Introduction
- 1.1Use the Language of Algebra
- 1.2Integers
- 1.3Fractions
- 1.4Decimals
- 1.5Properties of Real Numbers
- Chapter Review
  Key Terms
  Key Concepts
- Exercises
  Review Exercises
  Practice Test
2Solving Linear Equations
3Graphs and Functions
4Systems of Linear Equations
5Polynomials and Polynomial Functions
6Factoring
7Rational Expressions and Functions
8Roots and Radicals
9Quadratic Equations and Functions
10Exponential and Logarithmic Functions
11Conics
12Sequences, Series and Binomial Theorem
- Introduction
- 12.1Sequences
- 12.2Arithmetic Sequences
- 12.3Geometric Sequences and Series
- 12.4Binomial Theorem
- Chapter Review
  Key Terms
Answer Key
Index

Evaluate and Graph Exponential Functions

Learning Objectives

By the end of this section, you will be able to:

Graph exponential functions
Solve Exponential equations
Use exponential models in applications

Be Prepared 10.2

Before you get started, take this readiness quiz.

Simplify: $(\frac{x^{3}}{x^{2}}) . (\frac{x^{3}}{x^{2}}) .$
If you missed this problem, review Example 5.13.
Evaluate: ⓐ $2^{0} 2^{0}$ ⓑ ${(\frac{1}{3})}^{0} . {(\frac{1}{3})}^{0} .$
If you missed this problem, review Example 5.14.
Evaluate: ⓐ $2^{−1} 2^{−1}$ ⓑ ${(\frac{1}{3})}^{−1} . {(\frac{1}{3})}^{−1} .$
If you missed this problem, review Example 5.15.

Graph Exponential Functions

The functions we have studied so far do not give us a model for many naturally occurring phenomena. From the growth of populations and the spread of viruses to radioactive decay and compounding interest, the models are very different from what we have studied so far. These models involve exponential functions.

An exponential function is a function of the form $f (x) = a^{x} f (x) = a^{x}$ where $a > 0 a > 0$ and $a \neq 1 . a \neq 1 .$

Exponential Function

An exponential function, where $a > 0 a > 0$ and $a \neq 1, a \neq 1,$ is a function of the form

f (x) = a^{x} f (x) = a^{x}

Notice that in this function, the variable is the exponent. In our functions so far, the variables were the base.

This figure shows three functions: f of x equals negative 3x plus 4, which is marked as linear; f of x equals 2x squared plus 5x minus 3, which is marked as quadratic; and f of x equals 6 to the x power, which is marked exponential. For the functions marked linear and quadratic, x is the base. For the function marked exponential, x is the exponent for the base 6.

Our definition says $a \neq 1 . a \neq 1 .$ If we let $a = 1, a = 1,$ then $f (x) = a^{x} f (x) = a^{x}$ becomes $f (x) = 1^{x} . f (x) = 1^{x} .$ Since $1^{x} = 1 1^{x} = 1$ for all real numbers, $f (x) = 1^{} . f (x) = 1^{} .$ This is the constant function.

Our definition also says $a > 0 . a > 0 .$ If we let a base be negative, say $−4, −4,$ then $f (x) = {(−4)}^{x} f (x) = {(−4)}^{x}$ is not a real number when $x = \frac{1}{2} . x = \frac{1}{2} .$

\begin{matrix} f (x) & = & {(−4)}^{x} \\ f (\frac{1}{2}) & = & {(−4)}^{\frac{1}{2}} \\ f (\frac{1}{2}) & = & \sqrt{−4} not a real number \end{matrix} \begin{matrix} f (x) & = & {(−4)}^{x} \\ f (\frac{1}{2}) & = & {(−4)}^{\frac{1}{2}} \\ f (\frac{1}{2}) & = & \sqrt{−4} \end{matrix}

In fact, $f (x) = {(−4)}^{x} f (x) = {(−4)}^{x}$ would not be a real number any time $x x$ is a fraction with an even denominator. So our definition requires $a > 0 . a > 0 .$

By graphing a few exponential functions, we will be able to see their unique properties.

Example 10.10

On the same coordinate system graph $f (x) = 2^{x} f (x) = 2^{x}$ and $g (x) = 3^{x} . g (x) = 3^{x} .$

Show solution

Solution

We will use point plotting to graph the functions.

This table has seven rows and five columns. The first row is header row and reads x, f of x equals 2 to the x power, (x, f of x), g of x equals 3 to the x power, and (x, g of x). The second row reads negative 2, 2 to the negative 2 power equals 1 divided by 2 squared which equals 1 over 4, (negative 2, 1 over 4), 3 to the negative 2 power equals 1 divided by 3 squared which equals 1 over 9, (negative 2, 1 over 9). The third row reads negative 1, 2 to the negative 1 power equals 1 divided by 2 to the first power which equals 1 over 2, (negative 1, 1 over 2), 3 to the negative 1 power equals 1 divided by 3 to the first power which equals 1 over 3, (negative 1, 1 over 3). The fourth row reads 0, 2 to the 0 power equals 1, (0, 1), 3 to the 0 power equals 1, (0, 1). The fifth row reads 1, 2 to the 1 power equals 2, (1, 2), 3 to the 1 power equals 9, (1, 3). The sixth row reads 2, 2 to the 2 power equals 4, (2, 4), 3 to the 2 power equals 9, (2, 9). The seventh row reads 3, 2 to the 3 power equals 8, (3, 8), 3 to the 3 power equals 27, (3, 27).

This figure shows two curves. The first curve is marked in blue and passes through the points (negative 1, 1 over 2), (0, 1), and (1, 2). The second curve is marked in red and passes through the points (negative 1, 1 over 3), (0, 1), and (1, 3).

Try It 10.19

Graph: $f (x) = 4^{x} . f (x) = 4^{x} .$

Try It 10.20

Graph: $g (x) = 5^{x} . g (x) = 5^{x} .$

If we look at the graphs from the previous Example and Try Its, we can identify some of the properties of exponential functions.

The graphs of $f (x) = 2^{x} f (x) = 2^{x}$ and $g (x) = 3^{x}, g (x) = 3^{x},$ as well as the graphs of $f (x) = 4^{x} f (x) = 4^{x}$ and $g (x) = 5^{x}, g (x) = 5^{x},$ all have the same basic shape. This is the shape we expect from an exponential function where $a > 1 . a > 1 .$

We notice, that for each function, the graph contains the point $(0, 1) . (0, 1) .$ This make sense because $a^{0} = 1 a^{0} = 1$ for any a.

The graph of each function, $f (x) = a^{x} f (x) = a^{x}$ also contains the point $(1, a) . (1, a) .$ The graph of $f (x) = 2^{x} f (x) = 2^{x}$ contained $(1, 2) (1, 2)$ and the graph of $g (x) = 3^{x} g (x) = 3^{x}$ contained $(1, 3) . (1, 3) .$ This makes sense as $a^{1} = a . a^{1} = a .$

Notice too, the graph of each function $f (x) = a^{x} f (x) = a^{x}$ also contains the point $(−1, \frac{1}{a}) . (−1, \frac{1}{a}) .$ The graph of $f (x) = 2^{x} f (x) = 2^{x}$ contained $(−1, \frac{1}{2}) (−1, \frac{1}{2})$ and the graph of $g (x) = 3^{x} g (x) = 3^{x}$ contained $(−1, \frac{1}{3}) . (−1, \frac{1}{3}) .$ This makes sense as $a^{−1} = \frac{1}{a} . a^{−1} = \frac{1}{a} .$

What is the domain for each function? From the graphs we can see that the domain is the set of all real numbers. There is no restriction on the domain. We write the domain in interval notation as $(− \infty, \infty) . (− \infty, \infty) .$

Look at each graph. What is the range of the function? The graph never hits the $x x$ -axis. The range is all positive numbers. We write the range in interval notation as $(0, \infty) . (0, \infty) .$

Whenever a graph of a function approaches a line but never touches it, we call that line an asymptote. For the exponential functions we are looking at, the graph approaches the $x x$ -axis very closely but will never cross it, we call the line $y = 0, y = 0,$ the x-axis, a horizontal asymptote.

Properties of the Graph of $f (x) = a^{x} f (x) = a^{x}$ when $a > 1 a > 1$

Domain	$(− \infty, \infty) (− \infty, \infty)$
Range	$(0, \infty) (0, \infty)$
x-intercept	None
y-intercept	$(0, 1) (0, 1)$
Contains	$(1, a), (−1, \frac{1}{a}) (1, a),$
Asymptote	$x x$ -axis, the line $y = 0 y = 0$

This figure shows a curve that slopes upward from (negative 1, 1 over a) through (0, 1), up to (1, a).

Our definition of an exponential function $f (x) = a^{x} f (x) = a^{x}$ says $a > 0, a > 0,$ but the examples and discussion so far has been about functions where $a > 1 . a > 1 .$ What happens when $0 < a < 1 0 < a < 1$ ? The next example will explore this possibility.

Example 10.11

On the same coordinate system, graph $f (x) = {(\frac{1}{2})}^{x} f (x) = {(\frac{1}{2})}^{x}$ and $g (x) = {(\frac{1}{3})}^{x} . g (x) = {(\frac{1}{3})}^{x} .$

Show solution

Solution

We will use point plotting to graph the functions.

This table has seven rows and five columns. The first row is header row and reads x, f of x, equals 1 over 2 to the x power, (x, f of x), g of x equals 1 over 3 to the x power, and (x, g of x). The second row reads negative 2, 1 over 2 to the negative 2 power equals 2 squared which equals 4, (negative 2, 4), 3 to the negative 2 power equals 3 squared which equals 9, (negative 2, 9). The third row reads negative 1, 1 over 2 to the negative 1 power equals 2 to the first power which equals 2, (negative 1, 2), 1 over 3 to the negative 1 power equals 3 to the first power which equals 3, (negative 1, 3). The fourth row reads 0, 1 over 2 to the 0 power equals 1, (0, 1), 1 over 3 to the 0 power equals 1, (0, 1). The fifth row reads 1, 1 over 2 to the 1 power equals 1 over 2, (1, 1 over 2), 1 over 3 to the 1 power equals 1 over 3, (1, 1 over 3). The sixth row reads 2, 1 over 2 to the 2 power equals 1 over 4, (2, 1 over 4), 1 over 3 to the 2 power equals 1 over 9, (2, 1 over 9). The seventh row reads 3, 1 over 2 to the 3 power equals 1 over 8, (3, 1 over 8), 1 over 3 to the 3 power equals 1 over 27, (3, 1 over 27).

This figure shows two curves. The first curve is marked in blue and passes through the points (negative 1, 2), (0, 1), and (1, 1 over 2). The second curve is marked in red and passes through the points (negative 1, 3), (0, 1), and (1, 1 over 3).

Try It 10.21

Graph: $f (x) = {(\frac{1}{4})}^{x} . f (x) = {(\frac{1}{4})}^{x} .$

Try It 10.22

Graph: $g (x) = {(\frac{1}{5})}^{x} . g (x) = {(\frac{1}{5})}^{x} .$

Now let’s look at the graphs from the previous Example and Try Its so we can now identify some of the properties of exponential functions where $0 < a < 1 . 0 < a < 1 .$

The graphs of $f (x) = {(\frac{1}{2})}^{x} f (x) = {(\frac{1}{2})}^{x}$ and $g (x) = {(\frac{1}{3})}^{x} g (x) = {(\frac{1}{3})}^{x}$ as well as the graphs of $f (x) = {(\frac{1}{4})}^{x} f (x) = {(\frac{1}{4})}^{x}$ and $g (x) = {(\frac{1}{5})}^{x} g (x) = {(\frac{1}{5})}^{x}$ all have the same basic shape. While this is the shape we expect from an exponential function where $0 < a < 1, 0 < a < 1,$ the graphs go down from left to right while the previous graphs, when $a > 1, a > 1,$ went from up from left to right.

We notice that for each function, the graph still contains the point (0, 1). This make sense because $a^{0} = 1 a^{0} = 1$ for any a.

As before, the graph of each function, $f (x) = a^{x}, f (x) = a^{x},$ also contains the point $(1, a) . (1, a) .$ The graph of $f (x) = {(\frac{1}{2})}^{x} f (x) = {(\frac{1}{2})}^{x}$ contained $(1, \frac{1}{2}) (1, \frac{1}{2})$ and the graph of $g (x) = {(\frac{1}{3})}^{x} g (x) = {(\frac{1}{3})}^{x}$ contained $(1, \frac{1}{3}) . (1, \frac{1}{3}) .$ This makes sense as $a^{1} = a . a^{1} = a .$

Notice too that the graph of each function, $f (x) = a^{x}, f (x) = a^{x},$ also contains the point $(−1, \frac{1}{a}) . (−1, \frac{1}{a}) .$ The graph of $f (x) = {(\frac{1}{2})}^{x} f (x) = {(\frac{1}{2})}^{x}$ contained $(−1, 2) (−1, 2)$ and the graph of $g (x) = {(\frac{1}{3})}^{x} g (x) = {(\frac{1}{3})}^{x}$ contained $(−1, 3) . (−1, 3) .$ This makes sense as $a^{−1} = \frac{1}{a} . a^{−1} = \frac{1}{a} .$

What is the domain and range for each function? From the graphs we can see that the domain is the set of all real numbers and we write the domain in interval notation as $(− \infty, \infty) . (− \infty, \infty) .$ Again, the graph never hits the $x x$ -axis. The range is all positive numbers. We write the range in interval notation as $(0, \infty) . (0, \infty) .$

We will summarize these properties in the chart below. Which also include when $a > 1 . a > 1 .$

Properties of the Graph of $f (x) = a^{x} f (x) = a^{x}$

when $a > 1 a > 1$		when $0 < a < 1 0 < a < 1$
Domain	$(− \infty, \infty) (− \infty, \infty)$	Domain	$(− \infty, \infty) (− \infty, \infty)$
Range	$(0, \infty) (0, \infty)$	Range	$(0, \infty) (0, \infty)$
$x x$ -intercept	none	$x x$ -intercept	none
$y y$ -intercept	$(0, 1) (0, 1)$	$y y$ -intercept	$(0, 1) (0, 1)$
Contains	$(1, a), (−1, \frac{1}{a}) (1, a),$	Contains	$(1, a), (−1, \frac{1}{a}) (1, a),$
Asymptote	$x x$ -axis, the line $y = 0 y = 0$	Asymptote	$x x$ -axis, the line $y = 0 y = 0$
Basic shape	increasing	Basic shape	decreasing

This figure has two parts. On the left, we have a curve that passes through (negative 1, 1 over a) through (0, 1) to (1, a). On the right, where a is noted to be less than 1, we have a curve that passes through (negative 1, 1 over a) through (0, 1) to (1, a).

It is important for us to notice that both of these graphs are one-to-one, as they both pass the horizontal line test. This means the exponential function will have an inverse. We will look at this later.

When we graphed quadratic functions, we were able to graph using translation rather than just plotting points. Will that work in graphing exponential functions?

Example 10.12

On the same coordinate system graph $f (x) = 2^{x} f (x) = 2^{x}$ and $g (x) = 2^{x + 1} . g (x) = 2^{x + 1} .$

Show solution

Solution

We will use point plotting to graph the functions.

This table has seven rows and five columns. The first row is header row and reads x, f of x equals 2 to the x power, (x, f of x), g of x equals 2 to the x plus 1 power, and (x, g of x). The second row reads negative 2, 2 to the negative 2 power equals 1 divided by 2 squared which equals 1 over 4, (negative 2, 1 over 4), 2 to the negative 2 plus 1 power equals 1 divided by 2 to the first power which equals 1 over 2, (negative 2, 1 over 2). The third row reads negative 1, 2 to the negative 1 power equals 1 divided by 2 to the first power which equals 1 over 2, (negative 1, 1 over 2), 2 to the negative 1 plus 1 power equals 2 to the 0 power which equals 1, (negative 1, 1). The fourth row reads 0, 2 to the 0 power equals 1, (0, 1), 2 to the 0 plus 1 power equals 2 to the 1 power which equals 2, (0, 2). The fifth row reads 1, 2 to the 1 power equals 2, (1, 2), 2 to the 1 plus 1 power equals 2 to the second power which equals 4, (1, 4). The sixth row reads 2, 2 to the 2 power equals 4, (2, 4), 2 to the 2 plus 1 power equals 2 to the third power which equals 8, (2, 8). The seventh row reads 3, 2 to the 3 power equals 8, (3, 8), 2 to the 3 plus 1 power equals 2 to the fourth power which equals 16, (3, 16).

Try It 10.23

On the same coordinate system, graph: $f (x) = 2^{x} f (x) = 2^{x}$ and $g (x) = 2^{x - 1} . g (x) = 2^{x - 1} .$

Try It 10.24

On the same coordinate system, graph: $f (x) = 3^{x} f (x) = 3^{x}$ and $g (x) = 3^{x + 1} . g (x) = 3^{x + 1} .$

Looking at the graphs of the functions $f (x) = 2^{x} f (x) = 2^{x}$ and $g (x) = 2^{x + 1} g (x) = 2^{x + 1}$ in the last example, we see that adding one in the exponent caused a horizontal shift of one unit to the left. Recognizing this pattern allows us to graph other functions with the same pattern by translation.

Let’s now consider another situation that might be graphed more easily by translation, once we recognize the pattern.

Example 10.13

On the same coordinate system graph $f (x) = 3^{x} f (x) = 3^{x}$ and $g (x) = 3^{x} - 2 . g (x) = 3^{x} - 2 .$

Show solution

Solution

We will use point plotting to graph the functions.

This table has five rows and six columns. The first row is header row and reads x, f of x equals 3 to the x power, (x, f of x), g of x equals 3 to the x power minus 2, and (x, g of x). The second row reads negative 2, 3 to the negative 2 power equals 1 over 9, (negative 2, 1 over 9), 3 to the negative 2 power minus 2 equals 1 over 9 minus 2 which equals negative 17 over 9, (negative 2, negative 17 over 9). The third row reads negative 1, 3 to the negative 1 power equals 1 over 3, (negative 1, 1 over 3), 3 to the negative 1 power minus 2 equals 1 over 3 minus 2 which equals negative 5 over 3, (negative 1, negative 5 over 3). The fourth row reads 0, 3 to the 0 power equals 1, (0, 1), 3 to the 0 power minus 2 equals 1 minus 2 which equals negative 1, (0, negative 1). The fifth row reads 1, 3 to the 1 power equals 3, (1, 3), 3 to the 1 power minus 2 equals 3 minus 2 which equals 1, (1, 1). The sixth row reads 2, 3 squared equals 9, (2, 9), 3 squared minus 2 equals 9 minus 2 which equals 7, (2, 7).

This figure shows two curves. The first curve is marked in blue and passes through the points (negative 1, 1 over 3), (0, 1), and (1, 3). The second curve is marked in red and passes through the points (negative 1, negative 5 over 3), (0, negative 1), and (1, 1).

Try It 10.25

On the same coordinate system, graph: $f (x) = 3^{x} f (x) = 3^{x}$ and $g (x) = 3^{x} + 2 . g (x) = 3^{x} + 2 .$

Try It 10.26

On the same coordinate system, graph: $f (x) = 4^{x} f (x) = 4^{x}$ and $g (x) = 4^{x} - 2 . g (x) = 4^{x} - 2 .$

Looking at the graphs of the functions $f (x) = 3^{x} f (x) = 3^{x}$ and $g (x) = 3^{x} - 2 g (x) = 3^{x} - 2$ in the last example, we see that subtracting 2 caused a vertical shift of down two units. Notice that the horizontal asymptote also shifted down 2 units. Recognizing this pattern allows us to graph other functions with the same pattern by translation.

All of our exponential functions have had either an integer or a rational number as the base. We will now look at an exponential function with an irrational number as the base.

Before we can look at this exponential function, we need to define the irrational number, e. This number is used as a base in many applications in the sciences and business that are modeled by exponential functions. The number is defined as the value of ${(1 + \frac{1}{n})}^{n} {(1 + \frac{1}{n})}^{n}$ as n gets larger and larger. We say, as n approaches infinity, or increases without bound. The table shows the value of ${(1 + \frac{1}{n})}^{n} {(1 + \frac{1}{n})}^{n}$ for several values of $n . n .$

$n n$	${(1 + \frac{1}{n})}^{n} {(1 + \frac{1}{n})}^{n}$
1	2
2	2.25
5	2.48832
10	2.59374246
100	2.704813829…
1,000	2.716923932…
10,000	2.718145927…
100,000	2.718268237…
1,000,000	2.718280469…
1,000,000,000	2.718281827…

Table 10.1

e \approx 2.718281827 e \approx 2.718281827

The number e is like the number $π π$ in that we use a symbol to represent it because its decimal representation never stops or repeats. The irrational number e is called the natural base.

Natural Base $e e$

The number e is defined as the value of ${(1 + \frac{1}{n})}^{n}, {(1 + \frac{1}{n})}^{n},$ as n increases without bound. We say, as n approaches infinity,

e \approx 2.718281827.. . e \approx 2.718281827.. .

The exponential function whose base is $e, e,$ $f (x) = e^{x} f (x) = e^{x}$ is called the natural exponential function.

Natural Exponential Function

The natural exponential function is an exponential function whose base is $e e$

f (x) = e^{x} f (x) = e^{x}

The domain is $(− \infty, \infty) (− \infty, \infty)$ and the range is $(0, \infty) . (0, \infty) .$

Let’s graph the function $f (x) = e^{x} f (x) = e^{x}$ on the same coordinate system as $g (x) = 2^{x} g (x) = 2^{x}$ and $h (x) = 3^{x} . h (x) = 3^{x} .$

This figure shows the graphs of three functions. The first function, f of x equals 2 to the x, is marked in red and passes through the points (negative 1, negative 1 over 2), (0, negative 1), and (2, 1). The second function, f of x equals 3 to the x power, is marked in green and corresponds to a curve that passes through the points (negative 1, 1 over 3), (0, 1) and (1, 3). The third function, f of x equals e to the x power, is marked in blue and corresponds to a curve that passes through the points (negative 1, 1 over e), (0, 1) and (0, e).

Notice that the graph of $f (x) = e^{x} f (x) = e^{x}$ is “between” the graphs of $g (x) = 2^{x} g (x) = 2^{x}$ and $h (x) = 3^{x} . h (x) = 3^{x} .$ Does this make sense as $2 < e < 3 2 < e < 3$ ?

Solve Exponential Equations

Equations that include an exponential expression $a^{x} a^{x}$ are called exponential equations. To solve them we use a property that says as long as $a > 0 a > 0$ and $a \neq 1, a \neq 1,$ if $a^{x} = a^{y} a^{x} = a^{y}$ then it is true that $x = y . x = y .$ In other words, in an exponential equation, if the bases are equal then the exponents are equal.

One-to-One Property of Exponential Equations

For $a > 0 a > 0$ and $a \neq 1, a \neq 1,$

If a^{x} = a^{y}, then x = y . If

To use this property, we must be certain that both sides of the equation are written with the same base.

Example 10.14

How to Solve an Exponential Equation

Solve: $3^{2 x - 5} = 27 . 3^{2 x - 5} = 27 .$

Show solution

Solution

Step 1 is to write both sides of the equation with the same base. This means that, since the left side has base 3, we write the right side with base 3. Hence, 27 equals 3 to the third power. We have 3 to the 2x minus 5 power equals 27, which we write as 3 to the 2x minus 5 power equals 3 cubed.

Step 2 is to write a new equation by setting the exponents equal. This means that, since the bases are the same, the exponents must be equal. Hence, 2x minus 5 equals 3.

Step 3 is to solve the equation. This means that we add 5 to each side and divide by 2. Hence, 2x equals 8, which means that x equals 4.

Step 3 is to check the solution. This means that we substitute x equals 4 into the original equation. We start with 3 to the 2 x minus 5 power equals 27. We want to know whether 3 to the 2 times 4 minus 5 power equals 27. This becomes a question of whether 3 cubed equals 27, which of course is true.

Try It 10.27

Solve: $3^{3 x - 2} = 81 . 3^{3 x - 2} = 81 .$

Try It 10.28

Solve: $7^{x - 3} = 7 . 7^{x - 3} = 7 .$

The steps are summarized below.

How To

How to Solve an Exponential Equation

Step 1. Write both sides of the equation with the same base, if possible.
Step 2. Write a new equation by setting the exponents equal.
Step 3. Solve the equation.
Step 4. Check the solution.

In the next example, we will use our properties on exponents.

Example 10.15

Solve $\frac{e^{x^{2}}}{e^{3}} = e^{2 x} \frac{e^{x^{2}}}{e^{3}} = e^{2 x}$ .

Show solution

Solution

	$\frac{e^{x^{2}}}{e^{3}} = e^{2 x}$
Use the Property of Exponents: $\frac{a^{m}}{a^{n}} = a^{m - n} . \frac{a^{m}}{a^{n}} = a^{m - n} .$	$e^{x^{2} - 3} = e^{2 x}$
Write a new equation by setting the exponents equal.	$x^{2} - 3 = 2 x$
Solve the equation.	$x^{2} - 2 x - 3 = 0 x^{2} - 2 x - 3 = 0$
	$(x - 3) (x + 1) = 0 (x - 3) (x + 1) = 0$
	$x = 3, x = - 1 x = 3,$
Check the solutions.

Try It 10.29

Solve: $\frac{e^{x^{2}}}{e^{x}} = e^{2} . \frac{e^{x^{2}}}{e^{x}} = e^{2} .$

Try It 10.30

Solve: $\frac{e^{x^{2}}}{e^{x}} = e^{6} . \frac{e^{x^{2}}}{e^{x}} = e^{6} .$

Use Exponential Models in Applications

Exponential functions model many situations. If you own a bank account, you have experienced the use of an exponential function. There are two formulas that are used to determine the balance in the account when interest is earned. If a principal, P, is invested at an interest rate, r, for t years, the new balance, A, will depend on how often the interest is compounded. If the interest is compounded n times a year we use the formula $A = P {(1 + \frac{r}{n})}^{n t} . A = P {(1 + \frac{r}{n})}^{n t} .$ If the interest is compounded continuously, we use the formula $A = P e^{r t} . A = P e^{r t} .$ These are the formulas for compound interest.

Compound Interest

For a principal, P, invested at an interest rate, r, for t years, the new balance, A, is:

\begin{matrix} A = P {(1 + \frac{r}{n})}^{n t} & when compounded n times a year. \\ A = P e^{r t} & when compounded continuously. \end{matrix} \begin{matrix} A = P {(1 + \frac{r}{n})}^{n t} & when compounded & A = P e^{r t} & when compounded continuously. \end{matrix}

As you work with the Interest formulas, it is often helpful to identify the values of the variables first and then substitute them into the formula.

Example 10.16

A total of $$ 10,000 $ 10,000$ was invested in a college fund for a new grandchild. If the interest rate is $5 %, 5 %,$ how much will be in the account in 18 years by each method of compounding?

ⓐ compound quarterly

ⓑ compound monthly

Show solution

Solution

$\begin{matrix} \begin{matrix} Identify the values of each variable in the formulas. \\ Remember to express the percent as a decimal. \end{matrix} & \begin{matrix} A & = & ? \\ P & = & $ 10,000 \\ r & = & 0.05 \\ t & = & 18 years \end{matrix} \end{matrix} \begin{matrix} \begin{matrix} Identify the values of each variable in the formulas. \\ Remember to express the percent as a decimal. \end{matrix} & \begin{matrix} A & = & ? \\ P & = & $ 10,000 \\ r & = & 0.05 \\ t & = & 18 \end{matrix} \end{matrix}$

ⓐ
$\begin{matrix} \begin{matrix} For quarterly compounding, n = 4 . There are 4 \\ quarters in a year. \end{matrix} & A = P {(1 + \frac{r}{n})}^{n t} \\ Substitute the values in the formula. & A = 10,000 {(1 + \frac{0.05}{4})}^{4 \cdot 18} \\ \begin{matrix} Compute the amount. Be careful to consider the \\ order of operations as you enter the expression into \\ your calculator. \end{matrix} & A = $ 24,459.20 \end{matrix} \begin{matrix} \begin{matrix} For quarterly compounding, \\ quarters in a year. \end{matrix} \\ Substitute the values in the formula. \\ \begin{matrix} Compute the amount. Be careful to consider the \\ order of operations as you enter the expression into \\ your calculator. \end{matrix} \end{matrix}$

ⓑ
$\begin{matrix} \begin{matrix} For monthly compounding, n = 12 . There are 12 \\ months in a year. \end{matrix} & A = P {(1 + \frac{r}{n})}^{n t} \\ Substitute the values in the formula. & A = 10,000 {(1 + \frac{0.05}{12})}^{12 \cdot 18} \\ Compute the amount. & A = $ 24,550.08 \end{matrix} \begin{matrix} \begin{matrix} For monthly compounding, \\ months in a year. \end{matrix} \\ Substitute the values in the formula. \\ Compute the amount. \end{matrix}$

ⓒ
$\begin{matrix} For compounding continuously, & A = P e^{r t} \\ Substitute the values in the formula. & A = 10,000 e^{0.05 \cdot 18} \\ Compute the amount. & A = $ 24,596.03 \end{matrix} \begin{matrix} For compounding continuously, \\ Substitute the values in the formula. \\ Compute the amount. \end{matrix}$

Try It 10.31

Angela invested $$ 15,000 $ 15,000$ in a savings account. If the interest rate is $4 %, 4 %,$ how much will be in the account in 10 years by each method of compounding?

ⓐ compound quarterly

ⓑ compound monthly

Try It 10.32

Allan invested $10,000 in a mutual fund. If the interest rate is $5 %, 5 %,$ how much will be in the account in 15 years by each method of compounding?

ⓐ compound quarterly

ⓑ compound monthly

Other topics that are modeled by exponential functions involve growth and decay. Both also use the formula $A = P e^{r t} A = P e^{r t}$ we used for the growth of money. For growth and decay, generally we use $A_{0}, A_{0},$ as the original amount instead of calling it $P, P,$ the principal. We see that exponential growth has a positive rate of growth and exponential decay has a negative rate of growth.

Exponential Growth and Decay

For an original amount, $A_{0}, A_{0},$ that grows or decays at a rate, r, for a certain time, t, the final amount, A, is:

A = A_{0} e^{r t} A = A_{0} e^{r t}

Exponential growth is typically seen in the growth of populations of humans or animals or bacteria. Our next example looks at the growth of a virus.

Example 10.17

Chris is a researcher at the Center for Disease Control and Prevention and he is trying to understand the behavior of a new and dangerous virus. He starts his experiment with 100 of the virus that grows at a rate of 25% per hour. He will check on the virus in 24 hours. How many viruses will he find?

Show solution

Solution

$\begin{matrix} \begin{matrix} Identify the values of each variable in the formulas. \\ Be sure to put the percent in decimal form. \\ Be sure the units match—the rate is per hour and \\ the time is in hours. \end{matrix} & \begin{matrix} A & = & ? \\ A_{0} & = & 100 \\ r & = & 0.25 /hour \\ t & = & 24 hours \end{matrix} \\ Substitute the values in the formula: A = A_{0} e^{r t} . & A = 100 e^{0.25 \cdot 24} \\ Compute the amount. & A = 40,342.88 \\ Round to the nearest whole virus. & A = 40,343 \\ \begin{matrix} The researcher will find 40,343 \\ viruses. \end{matrix} \end{matrix} \begin{matrix} \begin{matrix} Identify the values of each variable in the formulas. \\ Be sure to put the percent in decimal form. \\ Be sure the units match—the rate is per hour and \\ the time is in hours. \end{matrix} & \begin{matrix} A & = & ? \\ A_{0} & = & 100 \\ r & = & 0.25 /hour \\ t & = & 24 \end{matrix} \\ Substitute the values in the formula: & A = 100 e^{0.25 \cdot 24} & Compute the amount. & A = 40,342.88 & Round to the nearest whole virus. & A = 40,343 & \begin{matrix} The researcher will find 40,343 \\ viruses. \end{matrix} \end{matrix}$

Try It 10.33

Another researcher at the Center for Disease Control and Prevention, Lisa, is studying the growth of a bacteria. She starts his experiment with 50 of the bacteria that grows at a rate of $15 % 15 %$ per hour. He will check on the bacteria every 8 hours. How many bacteria will he find in 8 hours?

Try It 10.34

Maria, a biologist is observing the growth pattern of a virus. She starts with 100 of the virus that grows at a rate of $10 % 10 %$ per hour. She will check on the virus in 24 hours. How many viruses will she find?

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

Practice Makes Perfect

Graph Exponential Functions

In the following exercises, graph each exponential function.

65.

$f (x) = 2^{x} f (x) = 2^{x}$

66.

$g (x) = 3^{x} g (x) = 3^{x}$

67.

$f (x) = 6^{x} f (x) = 6^{x}$

68.

$g (x) = 7^{x} g (x) = 7^{x}$

69.

$f (x) = {(1.5)}^{x} f (x) = {(1.5)}^{x}$

70.

$g (x) = {(2.5)}^{x} g (x) = {(2.5)}^{x}$

71.

$f (x) = {(\frac{1}{2})}^{x} f (x) = {(\frac{1}{2})}^{x}$

72.

$g (x) = {(\frac{1}{3})}^{x} g (x) = {(\frac{1}{3})}^{x}$

73.

$f (x) = {(\frac{1}{6})}^{x} f (x) = {(\frac{1}{6})}^{x}$

74.

$g (x) = {(\frac{1}{7})}^{x} g (x) = {(\frac{1}{7})}^{x}$

75.

$f (x) = {(0.4)}^{x} f (x) = {(0.4)}^{x}$

76.

$g (x) = {(0.6)}^{x} g (x) = {(0.6)}^{x}$

In the following exercises, graph each function in the same coordinate system.

77.

$f (x) = 4^{x}, g (x) = 4^{x - 1} f (x) = 4^{x},$

78.

$f (x) = 3^{x}, g (x) = 3^{x - 1} f (x) = 3^{x},$

79.

$f (x) = 2^{x}, g (x) = 2^{x - 2} f (x) = 2^{x},$

80.

$f (x) = 2^{x}, g (x) = 2^{x + 2} f (x) = 2^{x},$

81.

$f (x) = 3^{x}, g (x) = 3^{x} + 2 f (x) = 3^{x},$

82.

$f (x) = 4^{x}, g (x) = 4^{x} + 2 f (x) = 4^{x},$

83.

$f (x) = 2^{x}, g (x) = 2^{x} + 1 f (x) = 2^{x},$

84.

$f (x) = 2^{x}, g (x) = 2^{x} - 1 f (x) = 2^{x},$

In the following exercises, graph each exponential function.

85.

$f (x) = 3^{x + 2} f (x) = 3^{x + 2}$

86.

$f (x) = 3^{x - 2} f (x) = 3^{x - 2}$

87.

$f (x) = 2^{x} + 3 f (x) = 2^{x} + 3$

88.

$f (x) = 2^{x} - 3 f (x) = 2^{x} - 3$

89.

$f (x) = {(\frac{1}{2})}^{x - 4} f (x) = {(\frac{1}{2})}^{x - 4}$

90.

$f (x) = {(\frac{1}{2})}^{x} - 3 f (x) = {(\frac{1}{2})}^{x} - 3$

91.

$f (x) = e^{x} + 1 f (x) = e^{x} + 1$

92.

$f (x) = e^{x - 2} f (x) = e^{x - 2}$

93.

$f (x) = − 2^{x} f (x) = − 2^{x}$

94.

$f (x) = 2^{- x - 1} -1 f (x) = 2^{- x - 1} -1$

Solve Exponential Equations

In the following exercises, solve each equation.

95.

$2^{3 x - 8} = 16 2^{3 x - 8} = 16$

96.

$2^{2 x - 3} = 32 2^{2 x - 3} = 32$

97.

$3^{x + 3} = 9 3^{x + 3} = 9$

98.

$3^{x^{2}} = 81 3^{x^{2}} = 81$

99.

$4^{x^{2}} = 4 4^{x^{2}} = 4$

100.

$4^{x} = 32 4^{x} = 32$

101.

$4^{x + 2} = 64 4^{x + 2} = 64$

102.

$4^{x + 3} = 16 4^{x + 3} = 16$

103.

$2^{x^{2} + 2 x} = \frac{1}{2} 2^{x^{2} + 2 x} = \frac{1}{2}$

104.

$3^{x^{2} - 2 x} = \frac{1}{3} 3^{x^{2} - 2 x} = \frac{1}{3}$

105.

$e^{3 x} \cdot e^{4} = e^{10} e^{3 x} \cdot e^{4} = e^{10}$

106.

$e^{2 x} \cdot e^{3} = e^{9} e^{2 x} \cdot e^{3} = e^{9}$

107.

$\frac{e^{x^{2}}}{e^{2}} = e^{x} \frac{e^{x^{2}}}{e^{2}} = e^{x}$

108.

$\frac{e^{x^{2}}}{e^{3}} = e^{2 x} \frac{e^{x^{2}}}{e^{3}} = e^{2 x}$

In the following exercises, match the graphs to one of the following functions: ⓐ $2^{x} 2^{x}$ ⓑ $2^{x + 1} 2^{x + 1}$ ⓒ $2^{x - 1} 2^{x - 1}$ ⓓ $2^{x} + 2 2^{x} + 2$ ⓔ $2^{x} - 2 2^{x} - 2$ ⓕ $3^{x} 3^{x}$

109.

This figure shows an exponential that passes through (1, 1 over 3), (0, 1), and (1, 3).

110.

This figure shows an exponential that passes through (negative 2, 1 over 2), (negative 1, 1), and (0, 2).

111.

This figure shows an exponential that passes through (1, 1 over 2), (0, 1), and (1, 2).

112.

This figure shows an exponential that passes through (0, 1 over 2), (1, 1), and (2, 2).

113.

This figure shows an exponential that passes through (negative 1, 3 over 2), (0, negative 1), and (1, 0).

114.

This figure shows an exponential that passes through (negative 1, 5 over 2), (0, 3), and (1, 4).

Use exponential models in applications

In the following exercises, use an exponential model to solve.

115.

Edgar accumulated $$ 5,000 $ 5,000$ in credit card debt. If the interest rate is $20 % 20 %$ per year, and he does not make any payments for 2 years, how much will he owe on this debt in 2 years by each method of compounding? ⓐ compound quarterly ⓑ compound monthly ⓒ compound continuously

116.

Cynthia invested $$ 12,000 $ 12,000$ in a savings account. If the interest rate is $6 %, 6 %,$ how much will be in the account in 10 years by each method of compounding? ⓐ compound quarterly
ⓑ compound monthly ⓒ compound continuously

117.

Rochelle deposits $$ 5,000 $ 5,000$ in an IRA. What will be the value of her investment in 25 years if the investment is earning $8 % 8 %$ per year and is compounded continuously?

118.

Nazerhy deposits $$ 8,000 $ 8,000$ in a certificate of deposit. The annual interest rate is $6 % 6 %$ and the interest will be compounded quarterly. How much will the certificate be worth in 10 years?

119.

A researcher at the Center for Disease Control and Prevention is studying the growth of a bacteria. He starts his experiment with 100 of the bacteria that grows at a rate of $6 % 6 %$ per hour. He will check on the bacteria every 8 hours. How many bacteria will he find in 8 hours?

120.

A biologist is observing the growth pattern of a virus. She starts with 50 of the virus that grows at a rate of $20 % 20 %$ per hour. She will check on the virus in 24 hours. How many viruses will she find?

121.

In the last ten years the population of Indonesia has grown at a rate of $1.12 % 1.12 %$ per year to 258,316,051. If this rate continues, what will be the population in 10 more years?

122.

In the last ten years the population of Brazil has grown at a rate of $0.9 % 0.9 %$ per year to 205,823,665. If this rate continues, what will be the population in 10 more years?

Writing Exercises

123.

Explain how you can distinguish between exponential functions and polynomial functions.

124.

Compare and contrast the graphs of $y = x^{2} y = x^{2}$ and $y = 2^{x} y = 2^{x}$ .

125.

What happens to an exponential function as the values of $x x$ decreases? Will the graph ever cross the
$x x$ -axis? Explain.

Self Check

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

This table has four rows and four columns. The first row, which serves as a header, reads I can…, Confidently, With some help, and No—I don’t get it. The first column below the header row reads Graph exponential functions, solve exponential equations, and use exponential models in applications.

ⓑ After reviewing this checklist, what will you do to become confident for all objectives?

Learning Objectives

Be Prepared 10.2

Graph Exponential Functions

Exponential Function

Example 10.10

Solution

Try It 10.19

Try It 10.20

Properties of the Graph of f ( x ) = a x f ( x ) = a x when a > 1 a > 1

Example 10.11

Solution

Try It 10.21

Try It 10.22

Properties of the Graph of f ( x ) = a x f ( x ) = a x

Example 10.12

Solution

Try It 10.23

Try It 10.24

Example 10.13

Solution

Try It 10.25

Try It 10.26

Natural Base e e

Natural Exponential Function

Solve Exponential Equations

One-to-One Property of Exponential Equations

Example 10.14

How to Solve an Exponential Equation

Solution

Try It 10.27

Try It 10.28

How To

How to Solve an Exponential Equation

Example 10.15

Solution

Try It 10.29

Try It 10.30

Use Exponential Models in Applications

Compound Interest

Example 10.16

Solution

Try It 10.31

Try It 10.32

Exponential Growth and Decay

Example 10.17

Solution

Try It 10.33

Try It 10.34

Practice Makes Perfect

Writing Exercises

Self Check

Properties of the Graph of $f (x) = a^{x} f (x) = a^{x}$ when $a > 1 a > 1$

Properties of the Graph of $f (x) = a^{x} f (x) = a^{x}$

Natural Base $e e$