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Section 4.1 Exponential Functions

Before Starting this Section, Review

1 Integer exponents (Section P.2 , page 20)
2 Rational exponents (Section P.6 , page 61)
3 Graphing and transformations (Section 2.7 , page 263)
4 One-to-one functions (Section 2.9 , page 294)
5 Increasing and decreasing functions (Section 2.5 , page 235)

Objectives

1 Define an exponential function.
2 Graph exponential functions.
3 Develop formulas for simple and compound interest.
4 Understand the number e.
5 Define the natural exponential function.
6 Model using exponential functions.

Fooling a King

One night in northwest India, a wise man named Shashi invented a new game called “Shatranj” (chess). The next morning he took it to King Rai Bhalit, who was so impressed that he said to Shashi, “Name your reward.” Shashi merely requested that 1 grain of wheat be placed on the first square of the chessboard, 2 grains on the second, 4 on the third, 8 on the fourth, and so on, for all 64 squares. The king agreed to his request, thinking the man was an eccentric fool for asking for only a few grains of wheat when he could have had gold, jewels, or even his daughter’s hand in marriage.

You may know the end of this story. Much more wheat was needed to satisfy Shashi’s request. The king could never fulfill his promise to Shashi; instead, he had him beheaded.

The details of this rapidly growing wheat phenomenon are given in the margin. Table 4.1 shows the number of grains of wheat that need to be stacked on each square of the chessboard.

Table 4.1 Grains of Wheat on a Chessboard

Square Number	Grains Placed on this Square
1	1 (=20) $1 (= 2^{0})$
2	2 (=21) $2 (= 2^{1})$
3	4 (=22) $4 (= 2^{2})$
4	8 (=23) $8 (= 2^{3})$
5	16 (=24) $16 (= 2^{4})$
6	32 (=25) $32 (= 2^{5})$
⋅ $\cdot$	⋅ $\cdot$
⋅ $\cdot$	⋅ $\cdot$
⋅ $\cdot$	⋅ $\cdot$
63	262 $2^{62}$
64	263 $2^{63}$

These data can be modeled by the function

g (n) = 2 n - 1,

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

where g(n) is the number of grains of wheat and n is the number of the square (or the nth square) on the chessboard. Each square has twice the number of grains as the previous square. The function g is an example of an exponential function with base 2, with n restricted to 1, 2, 3, …, 64—the number of the chessboard squares. The name exponential function comes from the fact that the variable n occurs in the exponent. The base 2 is the growth factor.

When the growth rate of a quantity is directly proportional to the existing amount, the growth can be modeled by an exponential function. For example, exponential functions are often used to model the growth of investments and populations, the cell division of living organisms, and the decay of radioactive material. Example 12 discusses the growth of the world population.

Exponential Functions

1 Define an exponential function.

In the definition of the exponential function, we rule out the base a=1 $a = 1$ because in this case, the function is simply the constant function f(x)=1x, $f (x) = 1^{x},$ or f(x)=1. $f (x) = 1 .$ We exclude negative bases so that the domain includes all real numbers. For example, a cannot be −2 $- 2$ because (−2)1/2=−2−−−√ ${(- 2)}^{1 / 2} = \sqrt{- 2}$ is not a real number. Other functions with constant base and variables in the exponent, such as f(x)=4⋅3x, g(x)=−5⋅43x−2, $f (x) = 4 \cdot 3^{x}, g (x) = - 5 \cdot 4^{3 x - 2},$ and h(x)=c⋅ax $h (x) = c \cdot a^{x}$ (a>0 $a > 0$ and a≠1 $a \neq 1$ ), are also called exponential functions.

Evaluate Exponential Functions

We can evaluate exponential functions by using the laws of exponents and/or calculators, as in the next example. (See Section P.2 for the definitions and the laws of exponents involving ax $a^{x}$ when x is a rational number.)

Example 1 Evaluating Exponential Functions

Let f(x)=3x−2. $f (x) = 3^{x - 2} .$ Find f(4).
Let g(x)=−2⋅10x. $g (x) = - 2 \cdot 10^{x} .$ Find g(−2). $g (- 2) .$
Let h(x)=(19)x. $h (x) = {(\frac{1}{9})}^{x} .$ Find h(−32). $h (- \frac{3}{2}) .$
Let F(x)=4x. $F (x) = 4^{x} .$ Find F(3.2).

Solution

f(4)=34−2=32=9 $f (4) = 3^{4 - 2} = 3^{2} = 9$
g(−2)=−2⋅10−2=−2⋅1102=−2⋅1100=−0.02 $g (- 2) = - 2 \cdot 10^{- 2} = - 2 \cdot \frac{1}{10^{2}} = - 2 \cdot \frac{1}{100} = - 0.02$
h⎛⎝⎜⎜−32⎞⎠⎟⎟=(19)−32=(9−1)−32=932=(9–√)3=27 $h (- \frac{3}{2}) = {(\frac{1}{9})}^{- \frac{3}{2}} = {(9^{- 1})}^{- \frac{3}{2}} = 9^{\frac{3}{2}} = {(\sqrt{9})}^{3} = 27$
F(3.2)=43.2≈84.44850629Use a calculator. $F (3.2) = 4^{3.2} \approx 84.44850629 Use a calculator .$

Practice Problem 1

Let f(x)=(14)x. $f (x) = {(\frac{1}{4})}^{x} .$ Find f(2), f(0), f(−1), f(52), $f (2), f (0), f (- 1), f (\frac{5}{2}),$ and f(−32). $f (- \frac{3}{2}) .$

The domain of an exponential function is (−∞, ∞). $(- \infty, \infty) .$ In Example 1, we evaluated exponential functions at some rational numbers. But what is the meaning of ax $a^{x}$ when x is an irrational number? For example, what do expressions such as 32√ $3^{\sqrt{2}}$ and 2π $2^{π}$ mean? It turns out that the definition of ax $a^{x}$ (with a>0 $a > 0$ and x irrational) requires methods discussed in calculus. However, we can understand the basis for the definition from the following discussion. Suppose we want to define the number 2π. $2^{π} .$ We use several numbers that approximate π. $π .$ A calculator shows that π≈3.14159265…. $π \approx 3.14159265 \dots .$ We successively approximate 2π $2^{π}$ by using the rational powers shown in Table 4.2.

Table 4.2 Values of f(x)=2x $f (x) = 2^{x}$ for rational values of `x` that approach π $π$

`x`	3	3.1	3.14	3.141	3.1415
2x $2^{x}$	23=8 $2^{3} = 8$	23.1=8.5… $2^{3.1} = 8.5 \dots$	23.14=8.81… $2^{3.14} = 8.81 \dots$	23.141=8.821… $2^{3.141} = 8.821 \dots$	23.1415=8.8244… $2^{3.1415} = 8.8244 \dots$

It can be shown that the powers 23, 23.1, 23.14, 23.141, 23.1415,… $2^{3}, 2^{3.1}, 2^{3.14}, 2^{3.141}, 2^{3.1415}, \dots$ approach exactly one number. We define 2π $2^{π}$ as that number. Table 4.2 shows that 2π≈8.82 $2^{π} \approx 8.82$ (correct to two decimal places).

For our work with exponential functions, we need the following facts:

Exponential functions f(x)=ax $f (x) = a^{x}$ are defined for all real numbers x.
The graph of an exponential function is a continuous (unbroken) curve.
The rules of exponents hold for all real number exponents and positive bases.

Example 2 Using Rules of Exponents

Use the rules of exponents to simplify each expression.

627√⋅612√ $6^{\sqrt{27}} \cdot 6^{\sqrt{12}}$
(63√)3√ ${(6^{\sqrt{3}})}^{\sqrt{3}}$

Solution

627√⋅612√=====627√+12√633√+23√653√(65)3√77763√ax⋅ay=ax+y27−−√=9⋅3−−−−√=33–√; 12−−√=4⋅3−−−−√=23–√33–√+23–√=(3+2)3–√=53–√(ax)y=axy $\begin{array}{rcll} 6^{\sqrt{27}} \cdot 6^{\sqrt{12}} & = & 6^{\sqrt{27}}^{+ \sqrt{12}} & a^{x} \cdot a^{y} = a^{x + y} \\ = & 6^{3 \sqrt{3}}^{+ 2 \sqrt{3}} & \sqrt{27} = \sqrt{9 \cdot 3} = 3 \sqrt{3}; \sqrt{12} = \sqrt{4 \cdot 3} = 2 \sqrt{3} \\ = & 6^{5 \sqrt{3}} & 3 \sqrt{3} + 2 \sqrt{3} = (3 + 2) \sqrt{3} = 5 \sqrt{3} \\ = & {(6^{5})}^{\sqrt{3}} & {(a^{x})}^{y} = a^{x y} \\ = & 7776^{\sqrt{3}} \end{array}$
(63√)3√===63√⋅3√63216(ax)y=axy3–√⋅3–√=3⋅3−−−−√=9–√=3 $\begin{array}{rcll} {(6^{\sqrt{3}})}^{\sqrt{3}} & = & 6^{\sqrt{3}}^{\cdot \sqrt{3}} & {(a^{x})}^{y} = a^{x y} \\ = & 6^{3} & \sqrt{3} \cdot \sqrt{3} = \sqrt{3 \cdot 3} = \sqrt{9} = 3 \\ = & 216 \end{array}$

Practice Problem 2

Simplify each expression.
1. 38√⋅32√ $3^{\sqrt{8}} \cdot 3^{\sqrt{2}}$
2. (a8√)2√ ${(a^{\sqrt{8}})}^{\sqrt{2}}$

Graphing Exponential Functions

2 Graph exponential functions.

Let’s see how to sketch the graph of an exponential function. Although the domain is the set of all real numbers, we usually evaluate the functions only for the integer values of x (for ease of computation). To evaluate ax $a^{x}$ for noninteger values of x, use your calculator. Recall that the graph of a function f is the graph of the equation y=f(x). $y = f (x) .$ We can use either f(x)=ax $f (x) = a^{x}$ or y=ax $y = a^{x}$ to represent a given function.

Example 3 Graphing an Exponential Function

Graph the exponential function

f(x)=3x $f (x) = 3^{x}$
f(x)=(12)x. $f (x) = {(\frac{1}{2})}^{x} .$

Solution

First, make a table of a few values of x and the corresponding values of y.

`x`	−3 $- 3$	−2 $- 2$	−1 $- 1$	0	1	2	3
y=3x $y = 3^{x}$	127 $\frac{1}{27}$	19 $\frac{1}{9}$	13 $\frac{1}{3}$	1	3	9	27

Next, plot the points and draw a smooth curve through them. See Figure 4.1.

Figure 4.1

The graph of y=ax $y = a^{x}$ with a>1 $a > 1$ . Here a=3 $a = 3$

Note that the x-axis is the horizontal asymptote of the graph of y=ax. $y = a^{x} .$

Make a table similar to the one in part a.

`x`	−3 $- 3$	−2 $- 2$	−1 $- 1$	0	1	2	3
y=(12)x $y = {(\frac{1}{2})}^{x}$	8	4	2	1	12 $\frac{1}{2}$	14 $\frac{1}{4}$	18 $\frac{1}{8}$

Plotting these points and drawing a smooth curve through them, we get the graph of y=(12)x, $y = {(\frac{1}{2})}^{x},$ shown in Figure 4.2. In this graph, as x increases in the positive direction, y=(12)x $y = {(\frac{1}{2})}^{x}$ decreases toward 0. The x-axis is its horizontal asymptote.

Figure 4.2

An exponential function y=ax $y = a^{x}$ with 0<a<1 $0 < a < 1$ . Here a=12 $a = \frac{1}{2}$

Practice Problem 3

Sketch the graph of
1. f(x)=2x $f (x) = 2^{x}$
2. f(x)=(13)x. $f (x) = {(\frac{1}{3})}^{x} .$

In general, graphs of exponential functions have two basic shapes determined by the base a. If a>1, $a > 1,$ the graph of y=ax $y = a^{x}$ is rising as in Figure 4.1. If 0<a<1, $0 < a < 1,$ the graph is falling as in Figure 4.2.

Figure 4.3

Graphs of some exponential functions

Since y=(12)x=(2−1)x=2−x, $y = {(\frac{1}{2})}^{x} = {(2^{- 1})}^{x} = 2^{- x},$ the graph of y=(12)x $y = {(\frac{1}{2})}^{x}$ can also be obtained by reflecting the graph of y=2x $y = 2^{x}$ about the y-axis. See Figure 4.3.

In Figure 4.3, we sketch the graphs of four exponential functions on the same set of axes. These graphs illustrate some general properties of exponential functions.

Properties of Exponential Functions

Let y=f(x)=ax, a>0, a≠1. $y = f (x) = a^{x}, a > 0, a \neq 1 .$

The domain of f is (−∞, ∞). $(- \infty, \infty) .$
The range of f is (0, ∞): $(0, \infty) :$ The entire graph lies above the x-axis.
Note that f(x+1)=ax+1=a⋅ax=a f(x). $f (x + 1) = a^{x + 1} = a \cdot a^{x} = a f (x) .$ This means that the y-values change by a factor of a for each unit increase in x.
For a>1, $a > 1,$ the growth factor is a. Because the y-values increase by a factor of a for each unit increase in x,
1. f is an increasing function, so the graph rises to the right.
2. as x→∞, y→∞. $x \to \infty, y \to \infty .$
3. as x→−∞, y→0. $x \to - \infty, y \to 0 .$
For 0<a<1, $0 < a < 1,$ the decay factor is a. Because the y-values decrease by a factor of a for each unit increase in x,
1. f is a decreasing function, so the graph falls to the right.
2. as x→−∞, y→∞. $x \to - \infty, y \to \infty .$
3. as x→∞, y→0. $x \to \infty, y \to 0 .$
Each exponential function f is one-to-one. So,
1. if am=an, $a^{m} = a^{n},$ then m=n. $m = n .$
2. f has an inverse.
The graph of f has no x-intercepts, so it never crosses the x-axis. No value of x will cause f(x)=ax $f (x) = a^{x}$ to equal 0.
The graph of f is a smooth and continuous curve, and it passes through the points (−1, 1a), (0, 1), $(- 1, \frac{1}{a}), (0, 1),$ and (1, a).
The x-axis (y=0) $(y = 0)$ is a horizontal asymptote for the graph of every exponential function of the form f(x)=ax. $f (x) = a^{x} .$
The graph of y=a−x $y = a^{- x}$ is the reflection about the y-axis of the graph of y=ax. $y = a^{x} .$

Transformations on Exponential Functions

In Section 2.7, we discussed transformations on the graph of the function y=f(x) $y = f (x)$ . The transformations on the graph of y=ax $y = a^{x}$ are summarized in Table 4.3.

Table 4.3 Transformations on the graph of y=ax $y = a^{x}$

Transformation (in each case c>0 $c > 0$ )	Change the graph point (`x`, `y`) on y=ax $y = a^{x}$ to	Description
Vertical shift y=ax+c $y = a^{x} + c$ y=ax−c $y = a^{x} - c$	(x, y+c) $(x, y + c)$ (x, y−c) $(x, y - c)$	Shift the graph of y=ax $y = a^{x}$ up `c` units. Horizontal asymptote: y=c $y = c$ Shift the graph of y=ax $y = a^{x}$ down `c` units. Horizontal asymptote: y=−c $y = - c$
Horizontal shift y=ax−c $y = a^{x - c}$ y=ax+c $y = a^{x + c}$	(x+c, y) $(x + c, y)$ (x−c, y) $(x - c, y)$	Shift the graph of y=ax $y = a^{x}$ to the right `c` units. Shift the graph of y=ax $y = a^{x}$ to the left `c` units.
Reflection y=−ax $y = - a^{x}$ y=a−x $y = a^{- x}$	(x, −y) $(x, - y)$ (−x, y) $(- x, y)$	Reflect the graph of y=ax $y = a^{x}$ about the `x`-axis. Reflect the graph of y=ax $y = a^{x}$ about the `y`-axis.
Vertical stretching or compressing y=cax $y = c a^{x}$	(`x`, `cy`)	Vertically stretch the graph of y=ax $y = a^{x}$ if c>1. $c > 1.$ Vertically compress the graph of y=ax $y = a^{x}$ if 0<c<1. $0 < c < 1.$
Horizontal stretching or compressing y=acx $y = a^{c x}$	(xc, y) $(\frac{x}{c}, y)$	Horizontally compress the graph of y=ax $y = a^{x}$ if c>1. $c > 1.$ Horizontally stretch the graph of y=ax $y = a^{x}$ if 0<c<1. $0 < c < 1.$

Transformation (in each case

c>0 $c > 0$ )

Change the graph point (x, y) on

y=ax $y = a^{x}$ to

Description

Vertical shift

y=ax+c $y = a^{x} + c$

y=ax−c $y = a^{x} - c$

(x, y+c) $(x, y + c)$

(x, y−c) $(x, y - c)$

Shift the graph of y=ax $y = a^{x}$ up c units.

Horizontal asymptote: y=c $y = c$

Shift the graph of y=ax $y = a^{x}$ down c units.

Horizontal asymptote: y=−c $y = - c$

Horizontal shift

y=ax−c $y = a^{x - c}$

y=ax+c $y = a^{x + c}$

(x+c, y) $(x + c, y)$

(x−c, y) $(x - c, y)$

Shift the graph of y=ax $y = a^{x}$ to the right c units.

Shift the graph of y=ax $y = a^{x}$ to the left c units.

Reflection

y=−ax $y = - a^{x}$

y=a−x $y = a^{- x}$

(x, −y) $(x, - y)$

(−x, y) $(- x, y)$

Reflect the graph of y=ax $y = a^{x}$ about the x-axis.

Reflect the graph of y=ax $y = a^{x}$ about the y-axis.

Vertical stretching or compressing

y=cax $y = c a^{x}$

(x, cy)

Vertically stretch the graph of y=ax $y = a^{x}$ if c>1. $c > 1.$

Vertically compress the graph of y=ax $y = a^{x}$ if 0<c<1. $0 < c < 1.$

Horizontal stretching or compressing

y=acx $y = a^{c x}$

(xc, y) $(\frac{x}{c}, y)$

Horizontally compress the graph of y=ax $y = a^{x}$ if c>1. $c > 1.$

Horizontally stretch the graph of y=ax $y = a^{x}$ if 0<c<1. $0 < c < 1.$

To get the correct shape for the graph of f(x)=ax $f (x) = a^{x}$ , it is helpful to graph the three points (−1, 1a), (0, 1) $(- 1, \frac{1}{a}), (0, 1)$ , and (1, a). See Figure 4.4.

Figure 4.4

Basic shapes of y=ax $y = a^{x}$

To expedite the process of sketching the transformations on the graph of the exponential function f(x)=ax, $f (x) = a^{x},$ we focus on tracking the movement of the horizontal asmptote and the movement of the three points (−1, 1a), (0, 1), $(- 1, \frac{1}{a}), (0, 1),$ and (1, a) (see Figure 4.5 where points A, B, and C represent the final position of these three points, respectively, and the horizontal asymptote is shifted up). The final position of the horizontal asymptote will help to determine the range of the function produced by the sequence of transformations. If the final graph is above the horizontal asymptote y=k $y = k$ the range is (k, ∞), $(k, \infty),$ and if the final graph is below the horizontal asymptote y=k $y = k$ the range is (−∞, k). $(- \infty, k) .$ Note that horizontal shifts do not affect the location of the horizontal asymptote.

Figure 4.5

Characteristic features of the exponential functions

Example 4 Using Transformations on Exponential Functions

Start with the graph of y=2x $y = 2^{x}$ , and use transformations to sketch the graph of each function.

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

State the domain, range and the vertical asymptote for the graph of each function.

Solution

We start with the graph of f(x)=2x $f (x) = 2^{x}$ and use the transformations shown in Figure 4.6.

Figure 4.6

Transformations on y=2x $y = 2^{x}$

Practice Problem 4

Use the transformations of the graph y=2x $y = 2^{x}$ to sketch the graph of f(x)=−2x+3. $f (x) = - 2^{x} + 3 .$

Example 5 Using Transformations on Exponential Functions

Sketch the graph of f(x)=3−x+2 $f (x) = 3^{- x} + 2$ . Find the domain, range, and the horizontal asymptote of f.

Solution

Figure 4.7 shows a sequence of transformations on the graph of y=3x $y = 3^{x}$ used to produce the graph of f.

Figure 4.7

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

From the graph of f in Figure 4.7(c), we see that the domain of f is (−∞, ∞) $(- \infty, \infty)$ , its range is (2, ∞) $(2, \infty)$ , and the horizontal asymptote is the line y=2 $y = 2$ .

Practice Problem 5

Sketch the graph of f(x)=2x−1+3. $f (x) = 2^{x - 1} + 3 .$

Example 6 Finding Exponential Functions

Find the exponential function of the form f(x)=cax $f (x) = c a^{x}$ whose graph contains the points (−1, 18) $(- 1, 18)$ and (4, 227). $(4, \frac{2}{27}) .$

Solution

Write the function in the form y=f(x)=cax. $y = f (x) = c a^{x} .$

18 18 a 2 27 2 27 2 27 a 5 a = = = = = = = c a - 1 c c a 4 18 a a 4 18 a 5 1 243 1 243 - - - - \sqrt 5 = 1 3 (- 1, 18) is on the graph of y = c a x . Multiply both sides by a; a - 1 a = 1, simplify . (4, 2 27) is on the graph of y = c a x . Replace c with 18 a in the previous equation, a \cdot a 4 = a 5 Divide both sides by 18; simplify . Solve for a .

$\begin{array}{rcll} 18 & = & c a^{- 1} & (- 1, 18) is on the graph of y = c a^{x} . \\ 18 a & = & c & Multiply both sides by a; a^{- 1} a = 1, simplify . \\ \frac{2}{27} & = & c a^{4} & (4, \frac{2}{27}) is on the graph of y = c a^{x} . \\ \frac{2}{27} & = & 18 a a^{4} & Replace c with 18 a in the previous equation, \\ \frac{2}{27} & = & 18 a^{5} & a \cdot a^{4} = a^{5} \\ a^{5} & = & \frac{1}{243} & Divide both sides by 18; simplify . \\ a & = & \sqrt[5]{\frac{1}{243}} = \frac{1}{3} & Solve for a . \end{array}$

To find c, we substitute a=13 $a = \frac{1}{3}$ in the equation 18a=c. $18 a = c .$ So,

18 (1 3) = c or c = 6 .

$18 (\frac{1}{3}) = c or c = 6 .$

Replacing c=6 $c = 6$ and a=13 $a = \frac{1}{3}$ in f(x)=cax, $f (x) = c a^{x},$ we have:

f (x) = 6 (1 3) x .

$f (x) = 6 {(\frac{1}{3})}^{x} .$

Practice Problem 6

Repeat Example 6 with the graph points (−2, 16) $(- 2, 16)$ and (3, 12). $(3, \frac{1}{2}) .$

Simple Interest

3 Develop formulas for simple and compound interest.

We first review some terminology concerning interest.

When money is deposited with a bank, the bank becomes the borrower. For example, depositing $1000 in an account at 8% interest means that the principal P is $1000 and the interest rate r is 0.08 for the bank as the borrower.

Example 7 Calculating Simple Interest

Juanita has deposited $8000 in a bank for five years at a simple interest rate of 6%.

How much interest will she receive?
How much money will be in her account at the end of five years?

Solution

$\begin{array}{l} P = $ 8000, r = 0.06, and t = 5 . \\ \begin{array}{rcll} I & = & P r t & Equation (1) \\ = & $ 8000 (0.06) (5) & Substitute values . \\ = & $ 2400 & Simplify . \end{array} \end{array}$
In five years, the amount A she will receive is the principal plus the interest earned:

$\begin{array}{rcl} A & = & P + I \\ = & $ 8000 + $ 2400 \\ = & $ 10, 400 \end{array}$

Practice Problem 7

Find the amount that will be in a bank account if $10,000 is deposited at a simple interest rate of 7.5% for two years.

Given P and r, the amount A(t) due in t years and calculated at simple interest is found by using the formula

$A (t) = P + P r t .$ (2)

Equation (2) is a linear function of t. Simple interest problems are examples of linear growth, which take place when the growth of a quantity occurs at a constant rate and so can be modeled by a linear function.

Compound Interest

In the real world, simple interest is rarely used for periods of more than one year. Instead, we use compound interest—the interest paid on both the principal and the accrued (previously earned) interest.

To illustrate compound interest, suppose $1000 is deposited in a bank account paying 4% annual interest. At the end of one year, the account will contain the original $1000 plus the 4% interest earned on the $1000:

$$ 1000 + (0.04) ($ 1000) = $ 1040$

Similarly, if P represents the initial amount deposited at an interest rate r (expressed as a decimal) per year, then the amount $A_{1}$ in the account after one year is

$\begin{array}{rcll} A_{1} & = & P + r P & Principal P plus interest earned on P \\ = & P (1 + r) & Factor out P . \end{array}$

During the second year, the account earns interest on the new principal $A_{1} .$ The amount $A_{2}$ in the account after the second year will be equal to $A_{1}$ plus the interest on $A_{1} .$

$\begin{array}{rcll} A_{2} & = & A_{1} + r A_{1} & A_{1} plus interest earned on A_{1} \\ = & A_{1} (1 + r) & Factor out A_{1} . \\ = & P (1 + r) (1 + r) & A_{1} = P (1 + r) \\ = & P {(1 + r)}^{2} & (1 + r) (1 + r) = {(1 + r)}^{2} \end{array}$

The amount $A_{3}$ in the account after the third year is

$\begin{array}{rcll} A_{3} & = & A_{2} + r A_{2} & A_{2} plus interest earned on A_{2} \\ = & A_{2} (1 + r) & Factor out A_{2} . \\ = & P {(1 + r)}^{2} (1 + r) & A_{2} = P {(1 + r)}^{2} \\ = & P {(1 + r)}^{3} & {(1 + r)}^{2} (1 + r) = {(1 + r)}^{3} \end{array}$

In general, the amount A in the account after t years is given by

$A = P {(1 + r)}^{t} .$ (3)

We say that this type of interest is compounded annually because it is paid once a year.

Example 8 Calculating Compound Interest

Juanita deposits $8000 in a bank at the interest rate of 6% compounded annually for five years.

How much money will she have in her account after five years?
How much interest will she receive?

Solution

Here $P = $ 8000, r = 0.06, and t = 5; so$

$\begin{array}{rcll} A & = & P {(1 + r)}^{t} \\ = & $ 8000 {(1 + 0.06)}^{5} & P = $ 8000, r = 0.06, t = 5 \\ = & $ 8000 {(1.06)}^{5} \\ = & $ 10, 705.80 & Use a calculator . \end{array}$
$Interest = A - P = $ 10, 705.80 - $ 8000 = $ 2705.80$

Practice Problem 8

Repeat Example 8 assuming that the bank pays 7.5% interest compounded annually.

Comparing Examples 7 and 8, we see that compounding Juanita’s interest made her money grow faster. We expect this because the function $A (t) = P {(1 + r)}^{t}$ is an exponential function with base $(1 + r)$ that grows faster than a linear function with the same base.

Banks and other financial institutions usually pay savings account interest more than once a year. They pay a smaller amount of interest more frequently. Suppose the quoted annual interest rate r (also called the nominal rate) is compounded n times per year (at equal intervals) instead of annually. Then for each period, the interest rate is $\frac{r}{n},$ and there are $n \cdot t$ periods in t years. Accordingly, we can restate the formula $A (t) = P {(1 + r)}^{t}$ as follows:

The total amount accumulated after t years, denoted by A, is also called the future value of the investment.

Example 9 Using Different Compounding Periods to Compare Future Values

If $100 is deposited in a bank that pays 5% annual interest, find the future value A after one year if the interest is compounded

Annually.
Semiannually.
Quarterly.
Monthly.
Daily.

Solution

In the following computations, $P = $ 100, r = 0.05,$ and $t = 1 .$ Only n, the number of times interest is compounded each year, changes. Because $t = 1, n t = n (1) = n .$

Annual Compounding:

$\begin{array}{l} A = P {(1 + \frac{r}{n})}^{n t} n = 1; t = 1 \\ A = $ 100 (1 + 0.05) = $ 105.00 \end{array}$
Semiannual Compounding:

$\begin{array}{rcll} A & = & P {(1 + \frac{r}{2})}^{2} & n = 2; t = 1 \\ A & = & $ 100 {(1 + \frac{0.05}{2})}^{2} \\ \approx & $ 105.06 & Use a calculator . \end{array}$
Quarterly Compounding:

$\begin{array}{rcll} A & = & P {(1 + \frac{r}{4})}^{4} & n = 4; t = 1 \\ A & = & $ 100 {(1 + \frac{0.05}{4})}^{4} \\ \approx & $ 105.09 & Use a calculator . \end{array}$
Monthly Compounding:

$\begin{array}{rcll} A & = & P {(1 + \frac{r}{12})}^{12} & n = 12; t = 1 \\ A & = & $ 100 {(1 + \frac{0.05}{12})}^{12} \\ \approx & $ 105.12 & Use a calculator . \end{array}$
Daily Compounding:

$\begin{array}{rcll} A & = & P {(1 + \frac{r}{365})}^{365} & n = 365; t = 1 \\ A & = & $ 100 {(1 + \frac{0.05}{365})}^{365} \\ \approx & $ 105.13 & Use a calculator . \end{array}$

Practice Problem 9

Repeat Example 9 , assuming that $5000 is deposited at a 6.5% annual rate.

The next example shows that we can solve equation (4) for the interest rate r.

Side Note

You may make errors such as forgetting parentheses when entering complicated expressions in your calculator. Look at the answer to see whether it is reasonable and makes sense.

Example 10 Computing Interest Rate

Carmen has $9000 to invest. She needs $20,000 at the end of 8 years. If the interest is compounded quarterly, find the rate r needed.

Solution

Have $P = $ 9000, A = $ 20, 000, t = 8,$ and $n = 4 .$ Substituting these values in equation (4), we have

$$ 20, 000 = $ 9000 {(1 + \frac{r}{4})}^{4 \cdot 8}$

$\begin{array}{rcl} {(1 + \frac{r}{4})}^{32} & = & \frac{20, 000}{9000} & Rewrite, with 4 \cdot 8 = 32 and divide both sides by 9000. \\ = & \frac{20}{9} & Simplify the right side . \end{array}$

Taking the 32nd root or $\frac{1}{32}$ power of both sides, we have

$\begin{array}{rcll} 1 + \frac{r}{4} & = & {(\frac{20}{9})}^{1 / 32} \\ \frac{r}{4} & = & {(\frac{20}{9})}^{1 / 32} - 1 & Subtract 1 from both sides . \\ r & = & 4 [{(\frac{20}{9})}^{1 / 32} - 1] & Multiply both sides by 4. \\ \approx & 0.1010692264 & Use a calculator . \\ \approx & 10.107 % \end{array}$

Carmen needs an interest rate of about 10.107%.

Check: $9000 {(1 + \frac{0.10107}{4})}^{32} = 20, 000.12073 \approx 20, 000 .$

Practice Problem 10

Repeat Example 10 if the interest is compounded monthly.

Continuous Compound Interest Formula

4 Understand the number e.

Notice that in Example 9 the future value A increases with n, the number of compounding periods. (Of course P, r, and t are fixed.) The question is: If n increases indefinitely (100, 1000, 10,000 times, and so on), does the amount A also increase indefinitely? Let’s see why the answer is “no.” Let $h = \frac{n}{r} .$ Then we have

$\begin{array}{rcll} A & = & P {(1 + \frac{r}{n})}^{n t} & Equation (4) \\ = & P {[{(1 + \frac{r}{n})}^{n / r}]}^{r t} & n t = \frac{n}{r} \cdot r t \\ = & P {[{(1 + \frac{1}{h})}^{h}]}^{r t} & h = \frac{n}{r}, so \frac{1}{h} = \frac{r}{n} \end{array}$

Table 4.4 shows the expression ${(1 + \frac{1}{h})}^{h}$ as h takes on increasingly larger values.

Table 4.4

`h`	${(1 + \frac{1}{h})}^{h}$
1	2
2	2.25
10	2.59374
100	2.70481
1000	2.71692
10,000	2.71815
100,000	2.71827
1,000,000	2.71828

Leonhard Euler (1707–1783)

Leonhard Euler was the son of a Calvinist minister from Switzerland. At 13, Euler entered the University of Basel, pursuing a career in theology. At the university, Euler was tutored by Johann Bernoulli, of the famous Bernoulli family of mathematicians. Euler’s interest and skills led him to abandon his theological studies and take up mathematics. In 1741, Euler moved to the Berlin Academy, where he stayed until 1766. He then returned to St. Petersburg, where he remained for the rest of his life.

Euler contributed to many areas of mathematics, including number theory, combinatorics, and analysis, as well as applications to areas such as music and naval architecture. He wrote over 1100 books and papers and left so much unpublished work that it took 47 years after he died for all of his work to be published. During Euler’s life, his papers accumulated so quickly that he kept a large pile of articles awaiting publication. The Berlin Academy published the papers on top of this pile, so later results often were published before results they depended on or superseded. The project of publishing his collected works, undertaken by the Swiss Society of Natural Science, is still going on and will require more than 75 volumes.

Table 4.4 suggests that as h gets larger and larger, ${(1 + \frac{1}{h})}^{h}$ gets closer and closer to a fixed number. This observation can be proven, and the fixed number is denoted by e in honor of the famous mathematician Leonhard Euler (pronounced “oiler”). The number e, an irrational number, is sometimes called the Euler number.

We sometimes write

$e = \lim_{h \to \infty} {(1 + \frac{1}{h})}^{h},$

which means that when h is very large, ${(1 + \frac{1}{h})}^{h}$ has a value very close to e. Note that as n gets very large, the quantity $h = \frac{n}{r}$ also gets very large because r is fixed. Therefore, the compounded amount $A = P {(1 + \frac{r}{n})}^{n t} = P {[{(1 + \frac{r}{n})}^{n / r}]}^{r t}$ approaches $P e^{r t} .$

Continuous Compounding

When interest is compounded continuously, the amount A after t years is given by the following formula:

Example 11 Calculating Continuous Compound Interest

Find the amount when a principal of $8300 is invested at a 7.5% annual rate of interest compounded continuously for eight years and three months.

Solution

We use formula (5), with $P = $ 8300$ and $r = 0.075 .$ We convert eight years and three months to 8.25 years.

$\begin{array}{rcll} A & = & $ 8300 e^{(0.075) (8.25)} & Use the formula A = P e^{r t} . \\ \approx & $ 15, 409.83 & Use a calculator . \end{array}$

Practice Problem 11

Repeat Example 11 , assuming that $9000 is invested at a 6% annual rate.

The Natural Exponential Function

5 Define the natural exponential function.

The exponential function

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

with base e is so prevalent in the sciences that it is often referred to as the exponential function or the natural exponential function. We use a calculator to find $e^{x}$ to two decimal places for $x = - 2, - 1, 0, 1,$ and 2 in Table 4.5.

Table 4.5

`x`	$e^{x}$
$- 2$	0.14
$- 1$	0.37
0	1
1	2.72
2	7.39

The graph of $f (x) = e^{x}$ is sketched in Figure 4.8 by using the ordered pairs in Table 4.5.

Figure 4.8

The graph of $y = e^{x}$ is between the graphs $y = 2^{x}$ and $y = 3^{x}$

Because $2 < e < 3,$ the graph of $y = e^{x}$ lies between the graphs $y = 2^{x}$ and $y = 3^{x} .$ The function $f (x) = e^{x}$ has all the properties of exponential functions with base $a > 1$ listed on pages 426–427.

We can apply the transformations from Section 2.7 to the natural exponential function.

Exponential Growth and Decay

6 Model using exponential functions.

In Section 2.3 we discussed the properties of linear functions which are used to model linear growth and decay. Linear growth and decay occur when the growth (or decay) of a quantity is at a constant rate. Simple interest problems are examples of linear growth.

Side Note

In the unit intervals of time: Linear growth involves repeated addition

$C + C + C + C + C + \dots$

Exponential growth involves repeated multiplication

$C \cdot C \cdot C \cdot C \cdot C \cdot \dots$

If $k > 0$ , the model describes linear growth.
If $k < 0$ , the model describes linear decay.

Exponential functions are used to model exponential growth and decay. Exponential growth and decay occurs when the growth (or decay) of a quantity is directly proportional to the existing amount. Continuous compound interest problems are examples of exponential growth.

If $k > 0$ , the model describes exponential growth. Examples of exponential growth include microorganism growth, human population growth, and nuclear chain reaction.
If $k < 0$ , the model describes exponential decay. Examples of exponential decay include drug concentration, heat transfer, and radioactive decay.

Example 12 Exponential Growth

In the year 2000, the human population of the world was approximately 6.08 billion. Assume the annual rate of growth from 1990 onward at 1.5%. Using the exponential growth model, estimate the population of the world in the following years.

2030
1990

Solution

The year 2000 corresponds to $t = 0 .$ So $A_{0} = 6.08$ billion, $k = 0.015,$ and 2030 corresponds to $t = 30 .$

$\begin{array}{l} A (30) = 6.08 e^{(0.015) (30)} & A (t) = A_{0} e^{k t} \\ A (30) \approx 9.54 & Use a calculator . \end{array}$

Thus, the model predicts that if the rate of growth is 1.5% per year, over 9.54 billion people will be in the world in 2030.
The year 1990 corresponds to $t = - 10$ (because 1990 is ten years prior to 2000). We have

$\begin{array}{rcll} A (- 10) & = & 6.08 e^{(0.015) (- 10)} & A (t) = A_{0} e^{k t} \\ \approx & 5.23 & Use a calculator . \end{array}$

Thus, the model estimates that the world had over 5.23 billion people in 1990 assuming the growth rate had been 1.5% per year. (The actual population in 1990 was 5.28 billion.)

Practice Problem 12

Repeat Example 12 , assuming that the annual rate of growth is at 1.6%.

Example 13 Exponential Decay

You buy a fishing boat for $22,000. Your boat depreciates (exponentially) at the annual rate of 15% of its value. Find the depreciated value of your boat at the end of 5 years.

Solution

We use the exponential decay model: $A (t) = A_{0} \cdot e^{k t} (k < 0),$ where A(t) represents the value of the boat after t years.

Here $A_{0} = 22, 000, k = - 0.15,$ and $t = 5 .$ So

$\begin{array}{rcll} A (t) & = & 22, 000 e^{k t} \\ A (5) & = & 22, 000 e^{- (0.15) (5)} & Replace k with - 0.15 and t with 5. \\ \approx & 10, 392.06416 & Use a calculator . \end{array}$

So to the nearest cent, the value of your boat at the end of 5 years will be $10,392.06.

Practice Problem 13

Repeat Example 13 with $k = - 18 %$ and $t = 6$ years.

Section 4.1 Exercises

Concepts and Vocabulary

For the exponential function $f (x) = c a^{x}, a > 0, a \neq 1,$ the domain is , and for $c > 0$ the range is .
The graph of $f (x) = 3^{x}$ has y-intercept and has x-intercept.
The horizontal asymptote of the graph of $y = {(\frac{1}{3})}^{x}$ is the .
The exponential function $f (x) = a^{x}$ is increasing if and is decreasing if .
True or False. The graphs of $y = 2^{x}$ and $y = {(\frac{1}{2})}^{x}$ are symmetric with respect to the x-axis.
True or False. For the equation $y = a^{x} (a > 0, a \neq 1), y \to \infty$ as $x \to \infty .$
True or False. Let $y = e^{- x},$ then $y \to 0$ as $x \to \infty .$
True or False. The graph of $y = e^{x} + 1$ is obtained by shifting the graph of $y = e^{x}$ horizontally one unit to the left.

Building Skills

In Exercises 9–16, explain whether the given equation defines an exponential function. Give reasons for your answers. Write the base for each exponential function.

$y = x^{3}$
$y = 4^{x}$
$y = 2^{- x}$
$y = {(- 5)}^{x}$
$y = x^{x}$
$y = 4^{2}$
$y = 0^{x}$
$y = {(1.8)}^{x}$

In Exercises 17–22, evaluate each exponential function for the given value(s). (Use a calculator if necessary.)

$f (x) = 5^{x - 1}, f (0)$
$f (x) = - 2^{x + 1}, f (- 2)$
$g (x) = 3^{1 - x}; g (3.2), g (- 1.2)$
$g (x) = {(\frac{1}{2})}^{x + 1}; g (2.8), g (- 3.5)$
$h (x) = {(\frac{2}{3})}^{2 x - 1}; h (1.5), h (- 2.5)$
$f (x) = 3 - 5^{x}; f (2), f (- 1)$

In Exercises 23–30, simplify each expression. Write the answer in the form $a^{x} .$

$3^{\sqrt{2}} \cdot 3^{\sqrt{2}}$
$7^{\sqrt{3}} \cdot 49^{\sqrt{12}}$
$8^{π} \div 4^{π}$
$9^{\sqrt{5}} \div 27^{\sqrt{5}}$
${(3^{\sqrt{2}})}^{\sqrt{3}}$
${(2^{\sqrt{3}})}^{\sqrt{5}}$
${(a^{\sqrt{3}})}^{\sqrt{12}}$
$a^{\sqrt{2}} \cdot {(a^{2})}^{\sqrt{8}}$

In Exercises 31–34, find the function of the form $f (x) = c a^{x}$ that contains the two given graph points.

1. (0, 1) and (2, 16)
2. (0, 1) and $(- 2, \frac{1}{9})$
1. (0, 3) and (2, 12)
2. (0, 5) and (1, 15)
1. (1, 1) and (2, 5)
2. (1, 1) and $(2, \frac{1}{5})$
1. (1, 5) and (2, 125)
2. $(- 1, 4)$ and (1, 16)

In Exercises 35–42, sketch the graph of the given function by making a table of values. (Use a calculator if necessary.)

$f (x) = 4^{x}$
$g (x) = 10^{x}$
$g (x) = {(\frac{3}{2})}^{- x}$
$h (x) = 7^{- x}$
$h (x) = {(\frac{1}{4})}^{x}$
$f (x) = {(\frac{1}{10})}^{x}$
$f (x) = {(1.3)}^{- x}$
$g (x) = {(0.7)}^{- x}$

Match each exponential function given in Exercises 43–46 with one of the graphs labeled (a), (b), (c), and (d).

1. $f (x) = 5^{x}$
2. $f (x) = - 5^{x}$
3. $f (x) = 5^{- x}$
4. $f (x) = 5^{- x} + 1$

In Exercises 47–60, start with the graph of the appropriate basic exponential function f and use transformations to sketch the graph of the function g. State the domain and range of g and the horizontal asymptote of its graph.

$g (x) = 2^{x} + 1$
$g (x) = 3^{x} - 1$
$g (x) = 3^{x - 1}$
$g (x) = 2^{x + 1}$
$g (x) = 4^{- x}$
$g (x) = - 4^{x}$
$g (x) = 2^{x - 1} + 3$
$g (x) = 3^{x + 2} - 2$
$g (x) = - 3^{x - 2} + 1$
$g (x) = - 2^{x + 1} - 2$
$g (x) = {(\frac{1}{2})}^{x + 1} - 1$
$g (x) = {(\frac{1}{3})}^{x - 2} + 1$
$g (x) = - 2 \cdot 5^{x - 1} + 4$
$g (x) = \frac{1}{2} \cdot 5^{1 - x} - 2$

In Exercises 61–64, write an equation of the form $f (x) = a^{x} + b$ from the given graph. Then compute f(2).

In Exercises 65–68, write an equation of each graph in the final position.

The graph of $y = 2^{x}$ is shifted 2 units left and then 5 units up.
The graph of $y = 3^{x}$ is shifted 3 units right and then is reflected about the y-axis.
The graph of $y = {(\frac{1}{2})}^{x}$ is stretched vertically by a factor of 2 and then is shifted 5 units down.
The graph of $y = 2^{- x}$ is reflected about the x-axis and then is shifted 3 units up.

In Exercises 69 and 70, determine what sequence of transformations were applied on the graph of $f (x) = 2^{x}$ to produce the given graph.

In Exercises 71–74, find the simple interest for each value of principal P, rate r per year, and time t.

$P = $ 5000, r = 10 %, t = 5 years$
$P = $ 10, 000, r = 5 %, t = 10 years$
$P = $ 7800, r = 6 \frac{7}{8} %, t = 10 years$ and 9 months
$P = $ 8670, r = 4 \frac{1}{8} %, t = 6 years$ and 8 months

In Exercises 75–78, find (a) the future value of the given principal P and (b) the interest earned in the given period.

$P = $ 3500$ at 6.5% compounded annually for 13 years
$P = $ 6240$ at 7.5% compounded monthly for 12 years
$P = $ 7500$ at 5% compounded continuously for 10 years
$P = $ 8000$ at 6.5% compounded daily for 15 years

In Exercises 79–82, find the principal P that will generate the given future value A.

$A = $ 10, 000$ at 8% compounded annually for 10 years
$A = $ 10, 000$ at 8% compounded quarterly for 10 years
$A = $ 10, 000$ at 8% compounded daily for 10 years
$A = $ 10, 000$ at 8% compounded continuously for 10 years

In Exercises 83–90, starting with the graph of $y = e^{x},$ use transformations to sketch the graph of each function and state its horizontal asymptote.

$f (x) = e^{- x}$
$f (x) = - e^{x}$
$f (x) = e^{x - 2}$
$f (x) = e^{2 - x}$
$f (x) = 1 + e^{x}$
$f (x) = 2 - e^{- x}$
$f (x) = - e^{x - 2} + 3$
$g (x) = 3 + e^{2 - x}$

Applying the Concepts

Determine whether the quantity described below represents linear or exponential growth.
1. The population of a certain species of bird doubles every 10 years.
2. The number of people who reach Mt. Everest increases by 75 every year.
3. The number of visitors to a museum increases by 1.5% each year.
4. The population of a certain species of mammal increases by 100 every 4 years.
Determine whether the quantity described below represents linear or exponential decay.
1. The value of a certain household item depreciates by $100 every year.
2. A school of fish is losing one-fourth of its population every 2 years.
3. The concentration of a drug decreases by 5% every 10 minutes.
4. The tickets for a popular show are being sold at rate of 50 per day.
The cost of storage. The graph shows the average hard drive cost per one gigabyte during the years 2000–2014.

The function $f (t) = 7.89 e^{- 0.42 t}$ models these data, where t represents the number of years after 2000.
1. Determine whether the function represents exponential growth or decay.
2. Use this function to estimate the average hard drive cost per gigabyte in 2003
3. What trend does the graph follow?
Facebook revenues. The graph shows Facebook revenues in hundred millions of dollars during the years 2007–2014.

The function $f (t) = 1.92 e^{0.64 t}$ models these data, where t represents the number of years after 2007.
1. Determine whether the function represents exponential growth or decay.
2. Use this function to estimate Facebook revenue in 2015.
3. What trend does the graph follow?
Metal cooling. Suppose a metal block is cooling so that its temperature T (in $° C$ ) is given by $T = 200 \cdot 4^{- 0.1 t} + 25,$ where t is in hours.
1. Find the temperature after
  1. 2 hours.
  2. 3.5 hours.
2. How long has the cooling been taking place if the block now has a temperature of $125 ° C ?$
3. Find the eventual temperature $(t \to \infty) .$
Ethnic population. The population (in thousands) of people of East Indian origin in the United States is approximated by the function

$p (t) = 1600 {(2)}^{0.1047 t},$

where t is the number of years since 2010.
1. Find the population of this group in 2018.
2. Predict the population in 2025.
Price appreciation. In 2016, the median price of a house in Tampa was $190,000. Assuming a rate of increase of 3% per year, what can we expect the price of such a house to be in 2021?
Investment. How much should a mother invest at the time her son is born to provide him with $80,000 at age 21? Assume that the interest is 7% compounded quarterly.
Depreciation. Trans Trucking Co. purchased a truck for $80,000. The company depreciated the truck at the end of each year at the rate of 15% of its current value. What is the value of the truck at the end of the fifth year?
Manhattan Island purchase. In 1626, Peter Du Minuit purchased Manhattan Island from the Native Americans for 60 Dutch guilders (about $24). Suppose the $24 was invested in 1626 at a 6% rate. How much money would that investment be worth in 2006 if the interest was
1. Simple interest.
2. Compounded annually.
3. Compounded monthly.
4. Compounded continuously.
Investment. Ms. Ann Scheiber retired from government service in 1941 with a monthly pension of $83 and $5000 in savings. At the time of her death in January 1995 at the age of 101, Ms. Scheiber had turned the $5000 into $22 million through shrewd investments in the stock market. She bequeathed all of it to Yeshiva University in New York City. What annual rate of return compounded annually would turn $5000 to a whopping $22 million in 54 years?
Population. The population of Sometown, USA, was 12,000 in 2000 and grew to 15,000 in 2010. Assume that the population will continue to grow exponentially at the same constant rate. What will be the population of Sometown in 2020? $[H i n t : {Show that e}^{k} = {(\frac{5}{4})}^{1 / 10} .]$
Medicine. Tests show that a new ointment X helps heal wounds. If $A_{0}$ square millimeters is the area of the original wound, then the area A of the wound after n days of application of the ointment X is given by $A = A_{0} e^{- 0.43 n} .$ If the area of the original wound was 10 square millimeters, find the area of the wound after ten days of application of the new ointment.
Cooling. The temperature T (in $° C$ ) of coffee at time t minutes after its removal from the microwave is given by the equation

$T = 25 + 73 e^{- 0.28 t} .$

Find the temperature of the coffee at each time listed.
1. $t = 0$
2. $t = 10$
3. $t = 20$
4. after a long time
Interest rate. Mary purchases a 12-year bond for $60,000.00. At the end of 12 years, she will redeem the bond for approximately $95,600.00. If the interest is compounded quarterly, what was the interest rate on the bond?
Best deal. Fidelity Federal offers three types of investments: (i) 9.7% compounded annually, (ii) 9.6% compounded monthly, and (iii) 9.5% compounded continuously. Which investment is the best deal?
Balance transfer offers. Javier wants to transfer $1,000 from his high interest Credit Card account for the period of 12 months and needs to choose between two offers. Bank A is offering an 8% interest rate compounded monthly with a 2% initial transfer fee. Bank B is offering a 6% interest rate compounded monthly with a 6% initial transfer fee. Assume that in both cases the initial transfer fee in being added to the principal. Which bank provided Javier with a better offer and how much money he will save by choosing it over the other offer?
Paper stacking. Suppose we have a large sheet of paper 0.015 centimeter thick and we tear the paper in half and put the pieces on top of each other. We keep tearing and stacking in this manner, always tearing each piece in half. How high will the resulting pile of paper be if we continue the process of tearing and stacking
1. 30 times?
2. 40 times?
3. 50 times? [Hint: Use a calculator.]
Doubling. A jar with a volume of 1000 cubic centimeters contains bacteria that doubles in number every minute. If the jar is full in 60 minutes, how long did it take for the container to be half full?

Beyond the Basics

Let $f (x) = e^{x} .$ Show that
1. $\frac{f (x + h) - f (x)}{h} = e^{x} (\frac{e^{h} - 1}{h}) .$
2. $f (x + y) = f (x) f (y) .$
3. $f (- x) = \frac{1}{f (x)} .$
Let $f (x) = 3^{x} + 3^{- x}$ and $g (x) = 3^{x} - 3^{- x} .$ Find each of the following:
1. $f (x) + g (x)$
2. $f (x) - g (x)$
3. ${[f (x)]}^{2} - {[g (x)]}^{2}$
4. ${[f (x)]}^{2} + {[g (x)]}^{2}$
You need the following definition for this exercise: Let $0! = 1; 1! = 1; 2! = 2 \cdot 1;$ and, in general, $n! = n (n - 1) (n - 2) \cdot \dots \cdot 3 \cdot 2 \cdot 1 .$ The symbol n! is read “n factorial.” Now let

$S_{n} = 2 + \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} + \dots + \frac{1}{n!} .$

Compute $S_{n}$ for $n = 5, 10,$ and 15. Compare the resulting values with the value of e on page 435.
If an investment of P dollars returns A dollars after one year, the effective annual interest rate or annual yield y is defined by the equation

$A = P (1 + y) .$

Show that if the sum of P dollars is invested at a nominal rate r per year, compounded m times per year, the effective annual yield y is given by the equation

$y = {(1 + \frac{r}{m})}^{m} - 1 .$
The relative rate of growth of $A (t) = P e^{k t}$ in the interval $[t, t + h]$ is defined by $\frac{A (t + h) - A (t)}{A (t)} .$
1. Show that $\frac{A (t + 1) - A (t)}{A (t)} = e^{k} - 1 .$
2. Show that $\frac{A (t + h) - A (t)}{A (t)} = e^{k h} - 1 .$

In Exercises 115–118, describe an order in which a sequence of transformations is applied to the graph of $y = e^{x}$ to obtain the graph of the given equation.

$y = 3 e^{2 x - 1}$
$y = - 2 e^{3 x + 2}$
$y = 5 e^{2 - 3 x} + 4$
$y = - 2 e^{3 - 4 x} - 1$

Critical Thinking / Discussion / Writing

Discuss why we do not allow a to be 1, 0, or a negative number in the definition of an exponential function of the form $f (x) = a^{x} .$
Consider the function $g (x) = {(- 3)}^{x},$ which is not an exponential function. What are the possible rational numbers x for which g(x) is defined.
Discuss the end behavior of the graph of $y = \frac{b}{1 + c a^{x}},$ where $a, b, c > 0$ and $a \neq 1 .$ (Consider the cases $a < 1$ and $a > 1 .$ )
Write a summary of the types of graphs (with sketches) for the exponential functions of the form $y = c a^{x}, c \neq 0, a > 0,$ and $a \neq 1 .$
Give a convincing argument to show that the equation $2^{x} = k$ has exactly one solution for every $k > 0 .$ Support your argument with graphs.
Find all solutions of the equation $2^{x} = 2 x .$ How do you know you found all solutions?

Getting Ready for the Next Section

In Exercises 125–132, simplify each expression.

$10^{0}$
$10^{- 1}$
${(- 8)}^{1 / 3 \frac{}{}}$
${(25)}^{1 / 2}$
${(\frac{1}{7})}^{- 2}$
${(\frac{1}{9})}^{- 1 / 2}$
${(18)}^{1 / 2}$
${(\frac{1}{12})}^{- 1 / 2}$

In Exercises 133–138, find the inverse of each function. Then find the domain and the range of $f^{- 1} .$

$f (x) = 3 x + 4$
$f (x) = \frac{1}{2} x - 5$
$f (x) = \sqrt{x}, x \geq 0$
$f (x) = x^{2} + 1, x \geq 0$
$f (x) = \frac{1}{x - 1}$
$f (x) = \sqrt[3]{x}$

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Table of Contents for Section 4.1 Exponential Functions

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Table of Contents for
Section 4.1 Exponential Functions