Exponential Functions

Learning Objectives

Find the value of a function (exponential). (IA 3.5.3)
Graph exponential functions. (IA 10.2.1)

Objective 1: Find the value of a function (exponential). (IA 3.5.3)

Example 1

Evaluate the function $f (x) = 3^{x}$ for the given values

ⓐ $f (2)$
ⓑ $f (-1)$
ⓒ $f (2 h)$

Solution

ⓐ Replace x with 2 and find the value of the function $f (2) = 3^{2} = 9$
ⓑ Replace x with -1 and find the value of the function $f (2) = 3^{- 1} = \frac{1}{3}$
ⓒ Replace x with 2h and simplify if possible $f (2) = 3^{2 h}$

Practice Makes Perfect

Find the value of an exponential function.

Evaluate the function

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

for the given values.

ⓐ $f (2)$
ⓑ $f (-2)$
ⓒ $f (a)$

We also find the value of the function when we solve application problems involving exponential functions.
Medicare Premiums. The monthly Medicare Part B health-care premium for most beneficiaries ages 65 and older has increased significantly since 1975. The monthly premium has increased from about $7 in 1975 to $110.50 in 2011 (Source: Centers for Medicare and Medicaid Services). The following exponential function models the premium increases:
$M (x) = 7 {(1.080)}^{x}$ where x is the number of years since 1975.
Estimate the monthly Medicare Part B premium in 1985, in 1992, and in 2002. (Note that x is the number of years since 1975, so for 1985, x=10.) Round to the nearest dollar.

We can find Compound Interest using

A = P {(1 + \frac{r}{n})}^{n t}

,
Where A is the amount of money, P is the principal, t is the number of years, r is the interest rate, and n is the number of times the interest was compounded per year.
Suppose that $960 is invested at 7% interest, compounded semiannually.

ⓐ Find the function for the amount to which the investment grows after t years.
ⓑ Find the amount of money in the account at t=1, 6, 10, 15, and 20 years.

Objective 2: Graph exponential functions. (IA 10.2.1)

Practice Makes Perfect

Graph exponential functions.

Graph the exponential function

f (x) = 2^{x}

by making a table.

.
$x$	$y = f (x)$

A Cartesian coordinate system is shown with a grid. The x-axis ranges from -5 to 5, labeled at integer intervals. The y-axis ranges from -4 to 12, labeled at even integer intervals from -4 to 12. Both axes have arrows at their ends, and the labels 'x' and 'y' are present for their respective axes.

Graph the exponential function

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

by making a table.

.
$x$	$y = f (x)$

How does it compare with the graph of

f (x) = 2^{x}

Graph $f (x) = 3^{x}$ , $f (x) = 4^{x}$ , $f (x) = {2.5}^{x}$ in the same viewing window using a graphing calculator or program. What is the relationship between the base a and the shape of the graph?

Graph $f (x) = {0.2}^{x}$ , $f (x) = {0.4}^{x}$ , $f (x) = {0.7}^{x}$ in the same viewing window using a graphing calculator or program. What is the relationship between the base a and the shape of the graph?

Fill in the Properties of Exponential Function.
$f (x) = a^{x}, a > 0, a \neq 1$
Is it continuous?
Is it one-to-one?
Domain
Range
Increasing if
Decreasing if
Asymptotes
Intercepts

The number e, e ≈ 2.718281827, is like the number 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 or Euler's number after the Swiss mathematician Leonhard Euler.

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

Practice Makes Perfect

Graph the exponential function

f (x) = e^{x}

by making a table.

.
$x$	$y = f (x)$

What is the domain of

f (x)

?
What is the range of

f (x)

India is the second most populous country in the world with a population of about $1.39$ billion people in 2021. The population is growing at a rate of about $1.2 %$ each yearhttp://www.worldometers.info/world-population/. Accessed February 24, 2014.. If this rate continues, the population of India will exceed China’s population by the year $2027.$ 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 $f (x) = 3 x + 4,$ 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.

Table 1 Eight rows and three columns. The first column is labeled, “x”, which goes from 0 to 6; the second column is labeled, “f(x)=2^x”; and the third column is labeled, “g(x) = 2x”. The following values are for the function f: (0, 1), (1, 2), (2, 4), (3, 8), (4, 16), (5, 32), and (6, 64). The following values are for the function g: (0, 0), (1, 2), (2, 4), (3, 6), (4, 8), (5, 10), and (6, 12).
$x$	$f (x) = 2^{x}$	$g (x) = 2 x$
0	1	0
1	2	2
2	4	4
3	8	6
4	16	8
5	32	10
6	64	12

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 increasing by the same percentage over equal increments found in the domain.
Linear growth refers to the original value from the range increasing 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 $f (x) = a b^{x},$ where $a$ is any nonzero number, $b$ is a positive real number not equal to 1.

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

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

Table 2 Two rows and eight columns. The first row is labeled, “x”, and the second row is labeled, “f(x)=2^x”. Reading the columns as ordered pairs, we have the following values: (-3, 2^(-3)=1/8), (-2, 2^(-2)=1/4), (-1, 2^(-1)=1/2), (0, 2^(0)=1), (1, 2^(1)=2), (2, 2^(2)=4), and (3, 2^(3)=8).
$x$	$- 3$	$- 2$	$- 1$	$0$	$1$	$2$	$3$
$f (x) = 2^{x}$	$2^{- 3} = \frac{1}{8}$	$2^{- 2} = \frac{1}{4}$	$2^{- 1} = \frac{1}{2}$	$2^{0} = 1$	$2^{1} = 2$	$2^{2} = 4$	$2^{3} = 8$

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

Graph of Companies A and B’s functions, which values are found in the previous table. — Figure 1

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

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

Example 2

Identifying Exponential Functions

Which of the following equations are not exponential functions?

$f (x) = 4^{3 (x - 2)}$
$g (x) = x^{3}$
$h (x) = {(\frac{1}{3})}^{x}$
$j (x) = {(- 2)}^{x}$

Solution

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

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

Evaluating Exponential Functions

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

Let $b = - 9$ and $x = \frac{1}{2} .$ Then $f (x) = f (\frac{1}{2}) = {(- 9)}^{\frac{1}{2}} = \sqrt{- 9},$ which is not a real number.

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

Let $b = 1.$ Then $f (x) = 1^{x} = 1$ for any value of $x .$

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

Let $f (x) = 2^{x} .$ What is $f (3) ?$

\begin{array}{l} f (x) & = 2^{x} \\ f (3) & = 2^{3} & Substitute x = 3. \\ = 8 & Evaluate the power . \end{array}

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 $f (x) = 30 {(2)}^{x} .$ What is $f (3) ?$

\begin{array}{l} f (x) & = 30 {(2)}^{x} \\ f (3) & = 30 {(2)}^{3} & Substitute x = 3. \\ = 30 (8) & Simplify the power first . \\ = 240 & Multiply . \end{array}

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

f (3) = 30 {(2)}^{3} \neq 60^{3} = 216,000

Example 3

Evaluating Exponential Functions

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

Solution

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

\begin{array}{l} f (x) & = 5 {(3)}^{x + 1} \\ f (2) & = 5 {(3)}^{2 + 1} & Substitute x = 2. \\ = 5 {(3)}^{3} & Add the exponents . \\ = 5 (27) & Simplify the power . \\ = 135 & Multiply . \end{array}

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.

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 $A (x) = 100 + 50 x .$ 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 $B (x) = 100 {(1 + 0.5)}^{x} .$

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

Table 3 Six rows and three columns. The first column is labeled, “Year, x”, which goes from 0 to 3; the second column is labeled, “Stores, Company A”, which has a function of A(x) = 100+50x; and the third column is labeled, “Stores, Company B”, which has a function of B(x)=100(1+0.5)^x. The following values are for Company A’s function: (0, 100), (1, 150), (2, 200), and (3, 250). The following values are for the function Company B’s function: (0, 100), (1, 150), (2, 225), and (3, 337.5).
Year, $x$	Stores, Company A	Stores, Company B
$0$	$100 + 50 (0) = 100$	$100 {(1 + 0.5)}^{0} = 100$
$1$	$100 + 50 (1) = 150$	$100 {(1 + 0.5)}^{1} = 150$
$2$	$100 + 50 (2) = 200$	$100 {(1 + 0.5)}^{2} = 225$
$3$	$100 + 50 (3) = 250$	$100 {(1 + 0.5)}^{3} = 337.5$
$x$	$A (x) = 100 + 50 x$	$B (x) = 100 {(1 + 0.5)}^{x}$

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 $[0, \infty),$ and the range for both functions is $[100, \infty) .$ 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, $B (x) = 100 {(1 + 0.5)}^{x} .$ In this exponential function, 100 represents the initial number of stores, 0.50 represents the growth rate, and $1 + 0.5 = 1.5$ represents the growth factor. Generalizing further, we can write this function as $B (x) = 100 {(1.5)}^{x},$ where 100 is the initial value, $1.5$ is called the base, and $x$ is called the exponent.

Example 4

Evaluating a Real-World Exponential Model

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

Solution

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

P (18) = 1.25 {(1.012)}^{18} \approx 1.549

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

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 $a$ and $b,$ and evaluate the function.

Example 5

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 $N (t)$ representing the population $(N)$ of deer over time $t .$

Solution

We let our independent variable $t$ 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, $a = 80.$ We can now substitute the second point into the equation $N (t) = 80 b^{t}$ to find $b :$

\begin{array}{l} N (t) & = 80 b^{t} \\ 180 & = 80 b^{6} & Substitute using point (6, 180) . \\ \frac{9}{4} & = b^{6} & Divide and write in lowest terms . \\ b & = {(\frac{9}{4})}^{\frac{1}{6}} & Isolate b using properties of exponents . \\ b & \approx 1.1447 \begin{matrix}  \end{matrix} & Round to 4 decimal places . \end{array}

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 $N (t) = 80 {(1.1447)}^{t} .$ (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, $(0, 80)$ and $(6, 180) .$ We can also see that the domain for the function is $[0, \infty),$ and the range for the function is $[80, \infty) .$

Graph of the exponential function, N(t) = 80(1.1447)^t, with labeled points at (0, 80) and (6, 180). — Figure 3 Graph showing the population of deer over time, $N (t) = 80 {(1.1447)}^{t},$ $t$ years after 2006

Example 6

Writing an Exponential Model When the Initial Value is Not Known

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

Solution

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

Substituting $(- 2, 6)$ gives $6 = a b^{- 2}$
Substituting $(2, 1)$ gives $1 = a b^{2}$

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

Mathematical steps showing how to solve for 'a' from the equation 6 = ab^-2, by dividing by b^-2 and then using exponent properties to simplify to a = 6b^2.

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

The image shows steps on how to substitute the expression found for a into a second equation to find the value of b.

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

A mathematical equation shows 'a = 6b^2 ≈ 6(0.6389)^2 ≈ 2.4492', demonstrating the calculation and approximation of the variable 'a' based on a given value of 'b'.

Thus, the equation is $f (x) = 2.4492 {(0.6389)}^{x} .$

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, $(- 2, 6)$ and $(2, 1) .$ The graph is an example of an exponential decay function.

Graph of the exponential function, f(x)=2.4492(0.6389)^x, with labeled points at (-2, 6) and (2, 1). — Figure 4 The graph of $f (x) = 2.4492 {(0.6389)}^{x}$ models exponential decay.

Example 7

Writing an Exponential Function Given Its Graph

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

Graph of an increasing exponential function with notable points at (0, 3) and (2, 12). — Figure 5

Solution

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

\begin{array}{l} y = a b^{x} & Write the general form of an exponential equation . \\ y = 3 b^{x} & Substitute the initial value 3 for a . \\ 12 = 3 b^{2} & Substitute in 12 for y and 2 for x . \\ 4 = b^{2} & Divide by 3 . \\ b = \pm 2 \begin{matrix}  \end{matrix} & Take the square root . \end{array}

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

Example 8

Using a Graphing Calculator to Find an Exponential Function

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

Solution

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 $a = 6.2$ and $b = 2$ will be displayed. The exponential equation is $y = 6.2 \cdot 2^{x} .$

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 $t,$ principal $P,$ APR $r,$ and number of compounding periods in a year $n :$

A (t) = P {(1 + \frac{r}{n})}^{n t}

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.

Table 4 Six rows and two columns. The first column is labeled, “Frequency”, and the second column is labeled, “Value after 1 Year”. Reading the rows from left to right, we have that Annually is valued at 100, Semiannually at 102.50, Quarterly at 103.81, Monthly at 104.71, and Daily at 105.16.
Frequency	Value after 1 year
Annually	$1100
Semiannually	$1102.50
Quarterly	$1103.81
Monthly	$1104.71
Daily	$1105.16

Example 9

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?

Solution

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

\begin{array}{l} A (t) & = P {(1 + \frac{r}{n})}^{n t} & Use the compound interest formula . \\ A (10) & = 3000 {(1 + \frac{0.03}{4})}^{4⋅10} \begin{matrix}  \end{matrix} & Substitute using given values . \\ \approx $ 4045.05 & Round to two decimal places . \end{array}

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

Example 10

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?

Solution

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

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

\begin{array}{l} A (t) & = P {(1 + \frac{r}{n})}^{n t} & Use the compound interest formula . \\ 40,000 & = P {(1 + \frac{0.06}{2})}^{2 (18)} \begin{matrix}  \end{matrix} & Substitute using given values A, r, n, and t . \\ 40,000 & = P {(1.03)}^{36} & Simplify . \\ \frac{40,000}{{(1.03)}^{36}} & = P & Isolate P . \\ P & \approx $ 13, 801 & Divide and round to the nearest dollar . \end{array}

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

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.

Table 5 Nine rows and three columns. The first column is labeled, “Frequency”, the second column is labeled, “A(t)=(1+1/n)^x”, and the third column is labeled, “Value”. Reading the rows from left to right, we have that Annually has the input value of (1+1/1)^1 which equals to $2, and Semiannually has the input value of (1+1/2)^2 which equals to $2.25, Quarterly has the input value of (1+1/4)^4 which equals to $2.441406, Monthly has the input value of (1+1/12)^12 which equals to $2.613035, Daily has the input value of (1+1/365)^365 which equals to $2.714567, Hourly has the input value of (1+1/8766)^8766 which equals to $2.718127, One per minute has the input value of (1+1/525960)^525960 which equals to $2.718279, and Once per second has the input value of (1+1/31557600)^31557600 which equals to $2.718282.
Frequency	$A (n) = {(1 + \frac{1}{n})}^{n}$	Value
Annually	${(1 + \frac{1}{1})}^{1}$	$2
Semiannually	${(1 + \frac{1}{2})}^{2}$	$2.25
Quarterly	${(1 + \frac{1}{4})}^{4}$	$2.441406
Monthly	${(1 + \frac{1}{12})}^{12}$	$2.613035
Daily	${(1 + \frac{1}{365})}^{365}$	$2.714567
Hourly	${(1 + \frac{1}{8760})}^{8760}$	$2.718127
Once per minute	${(1 + \frac{1}{525600})}^{525600}$	$2.718279
Once per second	${(1 + \frac{1}{31536000})}^{31536000}$	$2.718282

These values appear to be approaching a limit as $n$ increases without bound. In fact, as $n$ gets larger and larger, the expression ${(1 + \frac{1}{n})}^{n}$ approaches a number used so frequently in mathematics that it has its own name: the letter $e .$ 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.

Example 11

Using a Calculator to Find Powers of e

Calculate $e^{3.14} .$ Round to five decimal places.

Solution

On a calculator, press the button labeled $[e^{x}] .$ The window shows $[e^(] .$ Type $3.14$ and then close parenthesis, $[)] .$ Press [ENTER]. Rounding to $5$ decimal places, $e^{3.14} \approx 23.10387.$ 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 $e .$

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 $e$ 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.

Example 12

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?

Solution

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

\begin{array}{l} A (t) & = P e^{r t} & Use the continuous compounding formula . \\ = 1000 {(e)}^{0.1} \begin{matrix}  \end{matrix} & Substitute known values for P, r, and t . \\ \approx 1105.17 & Use a calculator to approximate . \end{array}

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

Example 13

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?

Solution

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

\begin{array}{l} A (t) & = a e^{r t} & Use the continuous growth formula . \\ = 100 e^{- 0.173 (3)} \begin{matrix}  \end{matrix} & Substitute known values for a, r, and t . \\ \approx 59.5115 & Use a calculator to approximate . \end{array}

So 59.5115 mg of radon-222 will remain.

Key Equations

...
definition of the exponential function	$f (x) = b^{x}, where b > 0, b \neq 1$
definition of exponential growth	$f (x) = a b^{x}, where a > 0, b > 0, b \neq 1$
compound interest formula	$\begin{array}{l} A (t) = P {(1 + \frac{r}{n})}^{n t}, where \\ A (t) is the account value at time t \\ t is the number of years \\ P is the initial investment, often called the principal \\ r is the annual percentage rate (APR), or nominal rate \\ n is the number of compounding periods in one year \end{array}$
continuous growth formula	$A (t) = a e^{r t}, where$ $t$ is the number of unit time periods of growth $a$ is the starting amount (in the continuous compounding formula a is replaced with P, the principal) $e$ is the mathematical constant, $e \approx 2.718282$

Key Concepts

An exponential function is defined as a function with a positive constant other than $1$ raised to a variable exponent. See Example 2.
A function is evaluated by solving at a specific value. See Example 3 and Example 4.
An exponential model can be found when the growth rate and initial value are known. See Example 5.
An exponential model can be found when the two data points from the model are known. See Example 6.
An exponential model can be found using two data points from the graph of the model. See Example 7.
An exponential model can be found using two data points from the graph and a calculator. See Example 8.
The value of an account at any time $t$ can be calculated using the compound interest formula when the principal, annual interest rate, and compounding periods are known. See Example 9.
The initial investment of an account can be found using the compound interest formula when the value of the account, annual interest rate, compounding periods, and life span of the account are known. See Example 10.
The number $e$ is a mathematical constant often used as the base of real world exponential growth and decay models. Its decimal approximation is $e \approx 2.718282.$
Scientific and graphing calculators have the key $[e^{x}]$ or $[\exp (x)]$ for calculating powers of $e .$ See Example 11.
Continuous growth or decay models are exponential models that use $e$ as the base. Continuous growth and decay models can be found when the initial value and growth or decay rate are known. See Example 12 and Example 13.

Section Exercises

Verbal

Exercise 1

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

Solution

Linear functions have a constant rate of change. Exponential functions increase based on a percent of the original.

Exercise 2

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.

Exercise 3

The Oxford Dictionary defines the word nominal as a value that is “stated or expressed but not necessarily corresponding exactly to the real value.”Oxford Dictionary. http://oxforddictionaries.com/us/definition/american_english/nomina. 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.

Solution

When interest is compounded, the percentage of interest earned to principal ends up being greater than the annual percentage rate for the investment account. Thus, the annual percentage rate does not necessarily correspond to the real interest earned, which is the very definition of nominal.

Algebraic

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

Exercise 4

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

Exercise 5

A population of bacteria decreases by a factor of $\frac{1}{8}$ every $24$ hours.

Solution

exponential; the population decreases by a proportional rate. .

Exercise 6

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

Exercise 7

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

Solution

not exponential; the charge decreases by a constant amount each visit, so the statement represents a linear function. .

Exercise 8

The height of a projectile at time $t$ is represented by the function $h (t) = - 4.9 t^{2} + 18 t + 40.$

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

Exercise 9

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

Solution

The forest represented by the function $B (t) = 82 {(1.029)}^{t} .$

Exercise 10

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

Exercise 11

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

Solution

After $t = 20$ years, forest A will have $43$ more trees than forest B.

Exercise 12

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

Exercise 13

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?

Solution

Answers will vary. Sample response: For a number of years, the population of forest A will increasingly exceed forest B, but because forest B actually grows at a faster rate, the population will eventually become larger than forest A and will remain that way as long as the population growth models hold. Some factors that might influence the long-term validity of the exponential growth model are drought, an epidemic that culls the population, and other environmental and biological factors.

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

Exercise 14

$y = 300 {(1 - t)}^{5}$

Exercise 15

$y = 220 {(1.06)}^{x}$

Solution

exponential growth; The growth factor, $1.06,$ is greater than $1.$

Exercise 16

$y = 16.5 {(1.025)}^{\frac{1}{x}}$

Exercise 17

$y = 11, 701 {(0.97)}^{t}$

Solution

exponential decay; The decay factor, $0.97,$ is between $0$ and $1.$

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

Exercise 18

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

Exercise 19

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

Solution

$f (x) = 2000 {(0.1)}^{x}$

Exercise 20

$(- 1, \frac{3}{2})$ and $(3, 24)$

Exercise 21

$(- 2, 6)$ and $(3, 1)$

Solution

$f (x) = {(\frac{1}{6})}^{- \frac{3}{5}} {(\frac{1}{6})}^{\frac{x}{5}} \approx 2.93 {(0.699)}^{x}$

Exercise 22

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

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.

Exercise 23

$x$	1	2	3	4
$f (x)$	70	40	10	-20

Solution

Linear

Exercise 24

$x$	1	2	3	4
$h (x)$	70	49	34.3	24.01

Exercise 25

$x$	1	2	3	4
$m (x)$	80	61	42.9	25.61

Solution

Neither

Exercise 26

$x$	1	2	3	4
$f (x)$	10	20	40	80

Exercise 27

$x$	1	2	3	4
$g (x)$	-3.25	2	7.25	12.5

Solution

Linear

For the following exercises, use the compound interest formula, $A (t) = P {(1 + \frac{r}{n})}^{n t} .$

Exercise 28

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

Exercise 29

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

Solution

$$ 10, 250$

Exercise 30

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

Exercise 31

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

Solution

$$ 13, 268.58$

Exercise 32

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

Exercise 33

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

Solution

$P = A (t) \cdot {(1 + \frac{r}{n})}^{- n t}$

Exercise 34

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

Exercise 35

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

Solution

$$ 4,572.56$

Exercise 36

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

Exercise 37

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.

Solution

$4 %$

Exercise 38

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.

Exercise 39

$y = 3742 {(e)}^{0.75 t}$

Solution

continuous growth; the growth rate is greater than $0.$

Exercise 40

$y = 150 {(e)}^{\frac{3.25}{t}}$

Exercise 41

$y = 2.25 {(e)}^{- 2 t}$

Solution

continuous decay; the growth rate is less than $0.$

Exercise 42

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

Exercise 43

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

Solution

$$ 669.42$

Numeric

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

Exercise 44

$f (x) = 2 {(5)}^{x},$ for $f (- 3)$

Exercise 45

$f (x) = - 4^{2 x + 3},$ for $f (- 1)$

Solution

$f (- 1) = - 4$

Exercise 46

$f (x) = e^{x},$ for $f (3)$

Exercise 47

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

Solution

$f (- 1) \approx - 0.2707$

Exercise 48

$f (x) = 2.7 {(4)}^{- x + 1} + 1.5,$ for $f (- 2)$

Exercise 49

$f (x) = 1.2 e^{2 x} - 0.3,$ for $f (3)$

Solution

$f (3) \approx 483.8146$

Exercise 50

$f (x) = - \frac{3}{2} {(3)}^{- x} + \frac{3}{2},$ for $f (2)$

Technology

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

Exercise 51

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

Solution

$y = 3 \cdot 5^{x}$

Exercise 52

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

Exercise 53

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

Solution

$y \approx 18 \cdot {1.025}^{x}$

Exercise 54

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

Exercise 55

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

Solution

$y \approx 0.2 \cdot {1.95}^{x}$

Extensions

Exercise 56

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 $APY = {(1 + \frac{r}{12})}^{12} - 1.$

Exercise 57

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 $I (n)$ for the APY of any account that compounds $n$ times per year.

Solution

$APY = \frac{A (t) - a}{a} = \frac{a {(1 + \frac{r}{365})}^{365 (1)} - a}{a} = \frac{a [{(1 + \frac{r}{365})}^{365} - 1]}{a} = {(1 + \frac{r}{365})}^{365} - 1;$ $I (n) = {(1 + \frac{r}{n})}^{n} - 1$

Exercise 58

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

Exercise 59

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

Solution

Let $f$ be the exponential decay function $f (x) = a \cdot {(\frac{1}{b})}^{x}$ such that $b > 1.$ Then for some number $n > 0,$ $f (x) = a \cdot {(\frac{1}{b})}^{x} = a {(b^{- 1})}^{x} = a {({(e^{n})}^{- 1})}^{x} = a {(e^{- n})}^{x} = a {(e)}^{- n x} .$

Exercise 60

The formula for the amount $A$ in an investment account with a nominal interest rate $r$ at any time $t$ is given by $A (t) = a {(e)}^{r t},$ where $a$ 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 $t$ can be calculated with the formula $I (t) = e^{r t} - 1.$

Real-World Applications

Exercise 61

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?

Solution

$47, 622$ fox

Exercise 62

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?

Exercise 63

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?

Solution

$1.39 %;$ $$ 155, 368.09$

Exercise 64

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?

Exercise 65

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?

Solution

$$ 35, 838.76$

Exercise 66

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.)

Exercise 67

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?

Solution

$$ 82, 247.78;$ $$ 449.75$

Exercise 68

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.

annual percentage rate (APR)

the yearly interest rate earned by an investment account, also called nominal rate

compound interest

interest earned on the total balance, not just the principal

exponential growth

a model that grows by a rate proportional to the amount present

nominal rate

the yearly interest rate earned by an investment account, also called annual percentage rate