Financial Applications of Exponential Functions

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 $n$:

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

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

THE COMPOUND INTEREST FORMULA

Compound interest can be calculated using the formula

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

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.

EXAMPLE 8

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 $t=10$.

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

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

TRY IT #8

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

EXAMPLE 9

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 $k=2$. 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 $P$.

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

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

TRY IT #9

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

Source: Rice University, https://openstax.org/books/college-algebra/pages/6-1-exponential-functions
This work is licensed under a Creative Commons Attribution 4.0 License.

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

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.

THE NUMBER $e$

The letter $e$ represents the irrational number

$(1+ \frac{1}{n})^n$, as $n$ increases without bound

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

EXAMPLE 10

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 $\left[e^{x}\right]$. The window shows $\left[e^{\wedge}(]\right.$. 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 $e$.

TRY IT #10

Use a calculator to find $e^{−0.5}$. Round to five decimal places.

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.

THE CONTINUOUS GROWTH/DECAY FORMULA

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

$A(t)=ae^{rt}$

where

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

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

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

$A(t)=Pe^{rt}$

where

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

HOW TO

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

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

EXAMPLE 11

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:

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

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

TRY IT #11

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

EXAMPLE 12

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:

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

So 59.5115 mg of radon-222 will remain.

TRY IT #12

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