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# 3.6: Compound Interest

Difficulty Level: At Grade Created by: CK-12

## Learning objectives

• Calculate compound interest, including continuous compounding.
• Compare compound interest situations.
• Determine algebraically and graphically the time it takes for an account to reach a particular value.

## Introduction

In the previous lesson, you learned about modeling growth using an exponential function. In this lesson we will focus on a specific example of exponential growth: compounding of interest. We will begin with the case of simple interest, which refers to interest that is based only on the principal, or initial amount of an investment or loan. Then we will look at what it means for interest to compound. In the simplest terms, compounding means that interest accrues (you gain interest on an investment, or owe more on a loan) based on the principal you invested, as well as on interest you have already accrued. As you will soon see, compound interest is a case of exponential growth. In this lesson we will look at specific examples of compound interest, and we will write equations to model these specific situations.

## Simple interest over time

As noted above, simple interest means that interest accrues based on the principal of an investment or loan. The simple interest is calculated as a percent of the principal. The formula for simple interest is, in fact, simple:

The variable P represents the principal amount, r represents the interest rate, and t represents the amount of time the interest has been accruing. For example, say you borrow $2,000 from a family member, and you insist on repaying with interest. You agree to pay 5% interest, and to pay the money back in 3 years. The interest you will owe will be 2000(0.05)(3) =$300. This means that when you repay your loan, you will pay $2300. Note that the interest you pay after 3 years is not 5% of the original loan, but 15%, as you paid 5% of$2000 each year for 3 years.

Now let’s consider an example in which interest is compounded. Say that you invest $2000 in a bank account, and it earns 5% interest annually. How much is in the account after 3 years? In order to determine how much money is in the account after three years, we have to determine the amount of money in the account after each year. The table below shows the calculations for one, two, and three years of this investment: Year Principal + interest After one year 2000 + 2000(0.05) = 2000 + 100 =$2100
After 2 years 2100 + 2100(0.05) = 2100 + 105 = $2205 After 3 years 2205 + 2205(0.05) = 2205 + 110.25 =$2315.25

Therefore, after three years, you will have $2315.25 in the account, which means that you will have earned$315.25 in interest. With simple interest, you would have earned $300 in interest. Compounding results in more interest because the principal on which the interest is calculated increased each year. For example, in the second year shown in the table above, you earned 5% of 2100, not 5% of 2000, as would be the case of simple interest. The main idea here is that compounding creates more interest because you are earning interest on interest, and not just on the principal. But how much more? You might look at the above example and say, “it’s only$15.25.” Remember that we have only looked at one example, and this example is a small one: in the grand scheme of investing, $2000 is a small amount of money, and we have only looked at the growth of the investment for a short period of time. For example, if you are saving for retirement, you could invest for a period of 30 years or more, and you might invest several thousand dollars each year. The formulas and methods for calculating retirement investments are more complicated than what we will do here. However, we can use the above example to derive a formula that will allow us to calculate compound interest for any number of years. ## The compound interest formula To derive the formula for compound interest, we need to look at a more general example. Let’s return to the previous example, but instead of assuming the investment is$2000, let the principal of the investment be P dollars. The key idea is that each year you have 100% of the principal, plus 5% of the previous balance. The table below shows the calculations of this more general investment.

Year Principal + interest New principal
1 P + P × .05 = 1.00P + .05P = 1.05P
2 1.05P + .05 × (1.05P) = 1.05P [1+ .05] = 1.05P × 1.05 = (1.05)2 P
3 (1.05)2 P + .05 × (1.05)2 P = (1.05)2 P [1 + .05] = (1.05)2 P [1.05] = (1.05)3 P

Notice that at the end of every year, the amount of money in the investment is a power of 1.05, times P, and that the power corresponds to the number of years. Given this pattern, you might hypothesize that after 4 years, the amount of money is (1.05)4 P.

We can generalize this pattern to a formula. As above, we let P represent the principal of the investment. Now, let t represent the number of years, and r represent the interest rate. Keep in mind that 1.05 = 1+0.05. So we can generalize:

A(t) = P(1 + r)t

This function will allow us to calculate the amount of money in an investment, if the interest is compounded each year for t years.

Example 1: Use the formula above to determine the amount of money in an investment after 20 years, if you invest $2000, and the interest rate is 5% compounded annually. Solution: The investment will be worth$5306.60

A(t) = P(1 + r)t

A(20) = 2000(1.05)20

A(20) = $5306.60 In the above example, we found the value of this investment after a particular number of years. If we graph the function A(t) = 2000(1.05)t, we can see the values for any number of years. If you graph this function using a graphing calculator, you can determine the value of the investment by tracing along the function, or by pressing <TI font_TRACE> on your graphing calculator and then entering an x value. You can also choose an investment value you would like to reach, and then determine the number of years it would take to reach that amount. For example, how long will it take for the investment to reach$7,000?

As we did earlier in the chapter, we can find the intersection of the exponential function with the line y = 7000.

You can see from Figure 2 that the line and the curve intersect at a little less than x = 26. Therefore it would take almost 26 years for the investment to reach 7000. You can also solve for an exact value: \begin{align*}2000(1.05)^t = 7000\end{align*} \begin{align*}(1.05)^t = \frac{7000} {2000}\end{align*} Divide both sides by 2000 and simplify the right side of the equation. \begin{align*}(1.05)^t = 3.5\end{align*} Take the ln of both sides (you can use any log, but ln or log base 10 will allow you to use a calculator.) \begin{align*}ln (1.05)^t = ln 3.5\end{align*} \begin{align*}t[ ln (1.05)] = ln 3.5\end{align*} Use the power property of logs \begin{align*}t = \frac{ln 3.5} {ln 1.05} \approx 25.68\end{align*} Divide both sides by ln 1.05 The examples we have seen so far are examples of annual compounding. In reality, interest is often compounded more frequently, for example, on a monthly basis. In this case, the interest rate is divided amongst the 12 months. The formula for calculating the balance of the account is then slightly different: \begin{align*}A(t) = P \left (1 + \frac{r} {12}\right )^{12t}\end{align*} Notice that the interest rate is divided by 12 because 1/12th of the rate is applied each month. The variable t in the exponent is multiplied by 12 because the interest is calculated 12 times per year. In general, if interest is compounded n times per year, the formula is: \begin{align*}A(t) = P \left(1 + \frac{r} {n}\right )^{nt}\end{align*} Example 2: Determine the value of each investment. a. You invest5000 in an account that gives 6% interest, compounded monthly. How much money do you have after 10 years?
b. You invest $10,000 in an account that gives 2.5% interest, compounded quarterly. How much money do you have after 10 years? Solution: a.$5000, invested for 10 years at 6% interest, compounded monthly.
\begin{align*}A(t) = P \left (1 + \frac{r} {n}\right )^{nt}\end{align*}
\begin{align*}A(10) = 5000 \left (1 + \frac{.06} {12}\right )^{12\cdot 10}\end{align*}
\begin{align*}A(10) = 5000 \left (1.005\right )^{120}\end{align*}
\begin{align*}A(10) = \9096.98\end{align*}
b. 10000, invested for 10 years at 2.5% interest, compounded quarterly. Quarterly compounding means that interest is compounded four times per year. So in the equation, n = 4. \begin{align*}A(t) = P \left (1 + \frac{r} {n}\right )^{nt}\end{align*} \begin{align*}A(10) = 6000 \left( 1 + \frac{.025} {4}\right )^{4 \cdot 10}\end{align*} \begin{align*}A(10) = 6000 (1.00625)^{40}\end{align*} \begin{align*}A(10) = \12,830.30\end{align*} In each example, the value of the investment after 10 years depends on three quantities: the principal of the investment, the number of compoundings per year, and the interest rate. Next we will look at an example of one investment, but we will vary each of these quantities. ## The power of compound interest Consider the investment in example 1:2000 was invested at an annual interest rate of 5%. We modeled this situation with the equation A(t) = 2000(1.05)t. We can use this equation to determine the amount of money in the account after any number of years. As we saw above, the value of the account grows exponentially. You can see how fast the investment grows if we compare it to linear growth. For example, if the same investment earned simple interest, the value of the investment after t years could be modeled with the function B(t) = 2000 + 2000(.05)t. We can simplify this to be: B(t) = 100t + 2000. The exponential function and this linear function are shown here.

Notice that if we look at these investments over a long period of time (30 years are shown in the graph), the values look very close together for x values less than 10. For example, after 5 years, the compound interest investment is worth $2552.60, and the simple interest investment is worth$2500. But, after 20 years, the compound interest investment is worth $5306.60, and the simple interest investment is worth$4,000. After 20 years, simple interest has doubled the amount of money, while compound interest has resulted in 2.65 times the amount of money.

The main idea here is that an exponential function grows faster than a linear one, which you can see from the graphs above. But what happens to the investment if we change the interest rate, or the number of times we compound per year?

Example 3: Compare the values of the investments shown in the table. If everything else is held constant, how does the interest rate influence the value of the investment?

Principal r n t
a. $4,000 .02 12 8 b.$4,000 .05 12 8
c. $4,000 .10 12 8 d.$4,000 .15 12 8
e. 4,000 .22 12 8 Solution: Using the compound interest formula \begin{align*}A(t) = P\left (1 + \frac{r} {n}\right )^{nt}\end{align*}, we can calculate the value of each investment. In all cases, we have \begin{align*}A(8) = 4000 \left (1 + \frac{r} {12}\right )^{12 \cdot 8}\end{align*}. Principal r n t A a.4,000 .02 12 8 $4693.42 b.$4,000 .05 12 8 $5962.34 c.$4,000 .10 12 8 $8872.70$4,000 .15 12 8 $13182.05 e.$4,000 .22 12 8 22882.11 As we increase the interest rate, the value of the investment increases. It is part of every day life to want to find the highest interest rate possible for a bank account (and the lowest possible rate for a loan!). Let’s look at just how fast the value of the account grows. Remember that each calculation in the table above started with \begin{align*}A(8) = 4000 \left (1 + \frac{r} {12}\right )^{12 \cdot 8}\end{align*}. Notice that this is a function of r, the interest rate. We can write this equation in a more standard form: \begin{align*}f(x) = 4000 \left (1 + \frac{x} {12}\right )^{96}\end{align*}. The graph of this function is shown below: Notice that while this function is not exponential, it does grow quite fast. As we increase the interest rate, the value of the account increases very quickly. Example 4: Compare the values of the investments shown in the table. If everything else is held constant, how does the compounding influence the value of the investment? Principal r n t a.4,000 .05 1 (annual) 8
b. $4,000 .05 4 (quarterly) 8 c.$4,000 .05 12 (monthly) 8
d. $4,000 .05 365 (daily) 8 e.$4,000 .05 8760 (hourly) 8

Solution: Again, we use the compound interest formula. For this example, the n is the quantity that changes: \begin{align*}A(8) = 4000 \left (1 + \frac{.05} {n}\right )^{8n}\end{align*}

Principal r n t A
a. $4,000 .05 1 (annual) 8$5909.82
b. $4,000 .05 4 (quarterly) 8$5952.52
c. $4,000 .05 12 (monthly) 8$5962.34
d. $4,000 .05 365 (daily) 8$5967.14
e. $4,000 .05 8760 (hourly) 8$5967.29

In contrast to the changing interest rate, in this example, increasing the number of compoundings per year does not seem to dramatically increase the value of the investment. We can see why this is the case of we look at the function \begin{align*}A(8) = 4000 \left (1 + \frac{.05} {n}\right )^{8n}\end{align*}. A graph of the function \begin{align*}f(x) = 4000 \left (1 + \frac{.05} {x}\right )^{8x}\end{align*} is shown below:

The graph seems to indicate that the function has a horizontal asymptote at $6000. However, if we zoom in, we can see that the horizontal asymptote is closer to 5967. What does this mean? This means that for the investment of$4000, at 5% interest, for 8 years, compounding more and more frequently will never result in more than about 5968.00. Another way to say this is that the function \begin{align*}f(x) = 4000 \left (1 + \frac{.05} {x}\right )^{8x}\end{align*} has a limit as x approaches infinity. Next we will look at this kind of limit to define a special form of compounding. ## Continuous compounding Consider a hypothetical example: you invest1.00, at 100% interest, for 1 year. For this situation, the amount of money you have at the end of the year depends on how often the interest is compounded:

\begin{align*}A(t) = P \left (1 + \frac{r} {n}\right )^{nt}\end{align*}
\begin{align*}A(t) = 1 \left (1 + \frac{1} {n}\right )^{ln}\end{align*}
\begin{align*}A = \left (1 + \frac{1} {n}\right )^{n}\end{align*}

Now let’s consider different compoundings:

Compounding N A \begin{align*}\approx\end{align*}
Annual 1 2
Quarterly 4 2.44140625
Monthly 12 2.61303529022
Daily 365 2.71456748202
Hourly 8,760 2.71812669063
By the minute 525,600 2.7182792154
By the second 31,536,000 2.71828247254

The values of A in the table have a limit, which might look familiar: it’s the number e. In fact, one of the definitions of \begin{align*}e\end{align*} is \begin{align*}\lim_{n \to \infty} \left ( 1 + \frac{1} {n}\right )^n\end{align*}.

A related limit is one that will lead us to a special kind of compound interest: \begin{align*}\lim_{n \to \infty} \left (1 + \frac{x} {n}\right )^n = e^x\end{align*}. (The proof of this limit requires calculus. However, in one of the review questions, you will examine this limit more closely.)

Now we can define what is known as continuous compounding. If interest is compounded \begin{align*}n\end{align*} times per year, the equation we use is: \begin{align*}A(t) = P\left (1 + \frac{r} {n}\right )^{nt}\end{align*}. We can also write the function as \begin{align*}A(t) = P\left ((1 + \frac{r} {n})^n\right )^t\end{align*}. If we compound more and more often, we are looking at what happens to this function as \begin{align*}n \rightarrow \infty\end{align*}. Recall the limit above: \begin{align*}\lim_{n \to \infty} \left (1 + \frac{x} {n}\right )^n = e^x\end{align*}. Here, this means \begin{align*}\lim_{n \to \infty} \left (1 + \frac{r} {n}\right )^n = e^r\end{align*}. So as \begin{align*}n \end{align*} approaches \begin{align*}\infty\end{align*}, \begin{align*}\left ((1 + \frac{r} {n})^n\right )^t\end{align*} approaches \begin{align*} \,\! P(e^r)^t = Pe^{rt}\end{align*}.

The function A(t) = Pert is the formula we use to calculate the amount of money when interest is continuously compounded, rather than interest that is compounded at discrete intervals, such as monthly or quarterly. For example, consider again the investment in example 1 above: what is the value of an investment after 20 years, if you invest $2000, and the interest rate is 5% compounded continuously? A(t) = Pert A(20) = 2000e.05(20) A(20) = 2000e1 A(20) =$5436.56

Just as we did with the standard compound interest formula, we can also determine the time it takes for an account to reach a particular value if the interest is compounded continuously.

Example 5: How long will it take $2000 to grow to$25,000 in the previous example?

Solution: It will take about 50 years:

A(t) = Pert
25,000 = 2000e.05(t)
12.5 = e.05(t) Divide both sides by 2000
ln 12.5 = ln e.05(t) Take the ln of both sides
ln 12.5 = .05t ln e Use the power property of logs
ln 12.5 = .05t × 1 lne = 1
ln 12.5 = 0.5t Isolate t
\begin{align*}t = \frac{ln 12.5} {.05} \approx 50.5\end{align*}

## Lesson Summary

In this lesson we have developed formulas to calculate the amount of money in a bank account or an investment when interest is compounded, either a discrete number of times per year, or compounded continuously. We have found the value of accounts or investments, and we have found the time it takes to reach a particular value. We have solved these problems algebraically and graphically, using our knowledge of functions in general, and logarithms in particular.

In general, the examples we have seen are conservative in the larger scheme of investing. Given all of the information available today about investments, you may look at the examples and think that the return on these investments seems low. For example, in the last example, 50 years probably seems like a long time to wait!

It is important to keep in mind that these calculations are based on an initial investment only. In reality, if you invest money long term, you will invest on a regular basis. For example, if an employer offers a retirement plan, you may invest a set amount of money from every paycheck, and your employer may contribute a set amount as well. As noted above, the calculations for the growth of a retirement investment are more complicated. However, the exponential functions you have studied in this lesson are the basis for the calculations you would need to do. The examples here are meant to illustrate an application of exponential functions, and the power of compound interest.

## Points to Consider

1. Why is compound interest modeled with an exponential function?
2. What is the difference between compounding and continuous compounding?
3. How are logarithms useful in solving compound interest problems?

## Review Questions

1. You put $3500 in a bank account that earns 5.5% interest, compounded monthly. How much is in your account after 2 years? After 5 years? 2. You put$2000 in a bank account that earns 7% interest, compounded quarterly. How much is in your account after 10 years?
3. Solve an exponential equation in order to answer the question: given the investment in question 2, how many years will it take for the account to reach $10,000? 4. Use a graph to verify your answer to question 3. 5. Consider two investments: (1)$2000, invested at 6% interest, compounded monthly (2) $3000, invested at 4.5% interest, compounded monthly Use a graph to determine when the 2 investments have equal value. 6. You invest$3000 in an account that pays 6% interest, compounded monthly. How long does it take to double your investment?
7. Explain why the answer to #6 does not depend on the amount of the initial investment
8. You invest $4,000 in an account that pays 3.2% interest, compounded continuously. What is the value of the account after 5 years? 9. You invest$6,000 in an account that pays 5% interest, compounded continuously. What is the value of the account after 10 years?
10. Consider the investment in example 8. How many years will it take the investment to reach 20,000? 11. In this lesson, we introduced this limit: \begin{align*}\lim{n \to \infty} \left (1 + \frac{r} {n}\right )^n = e^r\end{align*} We noted in the lesson that the proof of this limit requires calculus. Here, we will examine a few specific cases in order to see how this limit is true. Using a graphing calculator, estimate the value of this limit for the given values of x. Do the limits seem to match the value of ex? Do these calculations suffice as a proof of the limit? Why or why not? Hint: Graph each limit expression as a function, where x represents n. 12. \begin{align*}r\end{align*} \begin{align*}\lim_{n \to \infty} \left (1 + \frac{r} {n}\right )^n = e^r\end{align*} \begin{align*}e^r\end{align*} 0 1 2 3 4 5 10 ## Review Answers 1. After 2 years:3905.99. After 5 years: $4604.96. 2.$4003.20
3. \begin{align*}t = \frac{ln 5} {4 ln 1.0175} \approx 23.19\end{align*} years.
4. The functions cross at x \begin{align*}\approx\end{align*} 23.19
5. It takes about 27 years for the two investments to have the same value.
6. \begin{align*}t = \frac{ln 2} {12 ln 1.005} \approx 11.58\end{align*} years.
7. When solving for t, the 6000 is divided by 3000, resulting in a 2 on the left side of the equation. (Hence the ln 2.) This would be the same, no matter what the initial investment was.
8. $4694.03 9.$9892.36
10. It will take about 50 years.
11. r \begin{align*}\lim_{n \to \infty} \left (1 + \frac{r} {n}\right )^n = e^r\end{align*} er
0 1 1
1 e e
2 \begin{align*}\approx\end{align*} 7.389 \begin{align*}\approx\end{align*} 7.389
3 \begin{align*}\approx\end{align*} 20.086 \begin{align*}\approx\end{align*} 20.086
4 \begin{align*}\approx\end{align*} 54.598 \begin{align*}\approx\end{align*} 54.598
5 \begin{align*}\approx\end{align*} 148.413 \begin{align*}\approx\end{align*} 148.413
10 \begin{align*}\approx\end{align*} 22026.466 \begin{align*}\approx\end{align*} 22026.466
12. The values in the table match, but this does not count as a proof. A proof needs to show that the values match for ALL values of
13. r
14. .

## Vocabulary

Accrue
To accrue is to increase in amount or value over time. If interest accrues on a bank account, you will have more money in your account. If interest accrues on a loan, you will owe more money to your lender.
Compound interest
Compound interest is interest based on a principal and on previous interest earned.
Continuous compounding
Interest that is based on continuous compounding is calculated according to the equation A(t)=Pert, where P is the principal, r is the interest rate, t is the length of the investment, and A is the value of the account or investment after t years.
Principal
The principal is the initial amount of an investment or a loan.
Simple interest
Simple interest is interest that is calculated as a percent of the principal, as a function of time.

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