Define and Evaluate Exponential Functions

Learning Outcomes

Identify and evaluate exponential functions
Use compound interest formulas

Linear functions have a constant rate of change – a constant number that the output increases for each increase in input. For example, in the equation $f(x)=3x+4$ , the slope tells us the output increases by three each time the input increases by one. Sometimes, on the other hand, quantities grow by a percent rate of change rather than by a fixed amount. In this lesson, we will define a function whose rate of change increases by a percent of the current value rather than a fixed quantity.

To illustrate this difference consider two companies whose business is expanding: Company A has $100$ stores and expands by opening $50$ new stores a year, while Company B has $100$ stores and expands by increasing the number of stores by $50\%$ of their total each year.

The table below compares the growth of each company where company A increases the number of stores linearly, and company B increases the number of stores by a rate of $50\%$ each year.

Year	Stores, Company A	Description of Growth	Stores, Company B
$0$	$100$	Starting with $100$ each	$100$
$1$	$100+50=150$	They both grow by $50$ stores in the first year.	$100$ $+50\%$ of $100$ $100 + 0.50(100) = 150$
$2$	$150+50=200$	Store A grows by $50$ , Store B grows by $75$	$150$ $+ 50\%$ of $150$ $150 + 0.50(150) = 225$
$3$	$200+50=250$	Store A grows by $50$ , Store B grows by $112.5$	$225 + 50\%$ of $225$ $225 + 0.50(225) = 337.5$

Company A has $100$ stores and expands by opening $50$ new stores a year, so its growth can be represented by the function $A\left(x\right)=100+50x$ . 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 $B\left(x\right)=100{\left(1+0.5\right)}^{x}$ .

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

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

The graph shows the number of stores Companies A and B opened over a five-year period.

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

Consider the function representing the number of stores for Company B:

$B\left(x\right)=100{\left(1+0.5\right)}^{x}$

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 $B\left(x\right)=100{\left(1.5\right)}^{x}$ , where $100$ is the initial value, $1.5$ is called the base, and x is called the exponent. This is an exponential function.

Exponential Growth

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

f (x) = a b^{x}

$f\left(x\right)=a{b}^{x}$

where

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

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

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

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

$\begin{array}{c}f\left(x\right)\hfill & ={2}^{x}\hfill & \hfill \\ f\left(3\right)\hfill & ={2}^{3}\text{ }\hfill & \text{Substitute }x=3.\hfill \\ \hfill & =8\text{ }\hfill & \text{Evaluate the power}\text{.}\hfill \end{array}$

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

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

$\begin{array}{c}f\left(x\right)\hfill & =30{\left(2\right)}^{x}\hfill & \hfill \\ f\left(3\right)\hfill & =30{\left(2\right)}^{3}\hfill & \text{Substitute }x=3.\hfill \\ \hfill & =30\left(8\right)\text{ }\hfill & \text{Simplify the power first}\text{.}\hfill \\ \hfill & =240\hfill & \text{Multiply}\text{.}\hfill \end{array}$

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

$f\left(3\right)=30{\left(2\right)}^{3}\ne {60}^{3}=216,000$

In our first example, we will evaluate an exponential function without the aid of a calculator.

Example

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

Show Solution

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

\begin{matrix} 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{matrix}

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

In the following video, we present more examples of evaluating an exponential function at several different values.

In the next example, we will revisit the population of India.

Example

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 $P\left(t\right)=1.25{\left(1.012\right)}^{t}$ , where t is the number of years since $2013$ . To the nearest thousandth, what will the population of India be in $2031$ ?

Show Solution

To estimate the population in $2031$ , we evaluate the model 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

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

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

In the following video, we show another example of using an exponential function to predict the population of a small town.

You may have seen formulas that are used to calculate compound interest rates. These formulas are another example of exponential growth. 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}

$A\left(t\right)=P{\left(1+\frac{r}{n}\right)}^{nt}$

The Compound Interest Formula

Compound interest can be calculated using the formula

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

$A\left(t\right)=P{\left(1+\frac{r}{n}\right)}^{nt}$

where

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

In our next example, we will calculate the value of an account after $10$ years of interest compounded quarterly.

Example

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

Show 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 $A(10)$ which is the value when t = $10$ .

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

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

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

The following video shows an example of using exponential growth to calculate interest compounded quarterly.

In our next example, we will use the compound interest formula to solve for the principal.

Example

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?

Show Solution

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

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{matrix} A (t) & = P {(1 + \frac{r}{n})}^{n t} & Use the compound interest formula . \\ 40, 000 & = P {(1 + \frac{0.06}{2})}^{2 (18)} & 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{matrix}

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

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

In the following video, we show another example of finding the deposit amount necessary to obtain a future value from compounded interest.

Summary

Exponential growth grows by a rate proportional to the current amount. For any real number x and any positive real numbers a and b such that $b\ne 1$ , an exponential growth function has the form $f\left(x\right)=a{b}^{x}$ . Evaluating exponential functions requires careful attention to the order of operations. Compound interest is an example of exponential growth.

Module 13: Exponential, and Logarithmic Functions