6.2 Compound Interest

Robert Foth

6.2 Compound Interest

6.2: Compound Interest

Learning Objectives

Upon completion of this section, you should be able to

Calculate the future value of an account for which interest is compounded $n$ times per year (discretely).
Solve application problems involving compounding interest discretely.
Calculate the future value of an account in applications for which interest is compounded continuously.
Solve application problems involving compounding interest continuously.

Compound Interest Discretely

In a standard bank account, any interest we earn is automatically added to our balance, and we earn interest on
that new balance. This reinvestment of interest is called compounding.

This compounding effect can lead to significantly larger growth (or debt accumulation) compared to simple interest, especially over longer periods of time. The frequency of compounding—whether daily, monthly, quarterly, or annually—also plays a crucial role in determining the final amount.

Compound Interest (discretely)

When interest is compounded a fixed number of times a year we can use the following formula to determine the future value of money:

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

$F V$ is the future value of the account after $t$ years.
$P$ is the starting balance of account (also called initial deposit, or principal).
$r$ is the annual interest rate in decimal form.
$n$ is the number of compounding periods (number of times per year that interest is calculated and added to the principal.)
$t$ is the length of time the money sits in the account in years.

One important thing to remember about using this formula is that it assumes that we put money in the account once and let it sit there earning interest.

The compounding period

n

is determined by how many times the interest is reinvested each year. Below are some common values we will use for

n

:

If the compounding is done annually (once a year), $n = 1$ .
If the compounding is done quarterly, $n = 4$ .
If the compounding is done monthly, $n = 12$ .
If the compounding is done weekly, $n = 52$ .
If the compounding is done daily, $n = 365$ .

To see how compound interest works lets take a look at this scenario: Suppose that we deposit $1000 in a bank account offering 3% interest, compounded monthly. How will our money grow each month over the year?

Since interest is being paid monthly, each month, we will earn interest at a rate of $\frac{3 %}{12} = 0.25 %$ per month. Notice that is what the $\frac{r}{n}$ represents in the formula (the interest paid per compounding period).

In the first month we can find the value in the account by finding the interest earned and adding that back into the principal:

$\begin{array}{l} P = $ 1000 \\ r = 0.0025 \\ I = $ 1000 (0.0025) = $ 2.50 \\ F V = $ 1000 + $ 2.50 = $ 1002.50 \end{array}$

After the first month we found that the account earned $2.50 in interest, which raising the account balance to $1002.50. In the second month the balance is now higher, so we calculate the interest on that new balance:

$\begin{array}{l} P = $ 1002.50 \\ r = 0.0025 \\ I = $ 1002.50 (0.0025) \approx $ 2.51 \\ F V = $ 1002.50 + $ 2.51 = $ 1005.01 \end{array}$

Notice that in the second month we earned more interest than we did in the first month (not much, but slightly more). This is because we earned interest not only on the original $1000 we deposited, but we also earned interest on the extra $2.50 (interest we earned the first month). This is the key advantage that compounding of interest gives us.

Calculating out a few more months:

Month	Starting balance	Interest earned	Ending Balance
1	1000.00	2.50	1002.50
2	1002.50	2.51	1005.01
3	1005.01	2.51	1007.52
4	1007.52	2.52	1010.04
5	1010.04	2.53	1012.57
6	1012.57	2.53	1015.10
7	1015.10	2.54	1017.64
8	1017.64	2.54	1020.18
9	1020.18	2.55	1022.73
10	1022.73	2.56	1025.29
11	1025.29	2.56	1027.85
12	1027.85	2.57	1030.42

In that table we can see how the amount of interest per month is increasing over time, which is something we do not see when working with simple interest as the principal amount will never change with simple interest calculations.

Let us compare the amount of money earned from compounding against the amount you would earn from an account that earned simple interest. Let us take an example where $3000 is deposited into an account that has an annual rate of 6%. Compare the future value of the account with simple interest to compound interest (monthly).

In the table below we can see the future value in the account in five year increments.

Years	Simple Interest ($15 per month)	6% compounded monthly = 0.5% each month.
0	$3000	$3000
5	$3900	$4046.55
10	$4800	$5458.19
15	$5700	$7362.28
20	$6600	$9930.61
25	$7500	$13394.91
30	$8400	$18067.73
35	$9300	$24370.65

Graph of Simple vs Compound Interest Go to Desmos to explore this further.

As you can see, over a long period of time, compounding makes a large difference in the account balance. You may recognize the graphs as the difference between linear growth and exponential growth.

Example 1

A certificate of deposit (CD) is savings instrument that many banks offer. It usually gives a higher interest rate, but you cannot access your investment for a specified length of time. Suppose you deposit $3000 in a CD paying 6% interest, compounded monthly. How much will you have in the account after 20 years?

Solution

Let us start with listing out the given information.

$\begin{array}{l} P = $ 3000 & the initial deposit \\ r = 0.06 & 6 % annual rate \\ n = 12 & 12 months in 1 year \\ t = 20 & since we ’ re looking for how much we ’ ll have after 20 years \end{array}$

Next we can use the formula to find the future value:

$FV = 3000 {(1 + \frac{0.06}{12})}^{20 \times 12} = $ 9930.61$ (round your answer to the nearest penny)

As you enter this expression into a calculator you will want to ensure that the exponent term is in a parenthesis to ensure you are evaluating the entire product of 20 and 12 in the exponent. It should look something like this: 3000*(1+0.06/12)^(20*12)

Try it Now 1

Peter deposits $200 that was given to him upon graduation into an account with an annual rate of 7% compounded weekly. What is the value of that deposit after 4 years? After 8 years?

Hint 1 (click to Show/Hide)

We are compounding the interest, so you will need to use the compound interest formula. The compound interest formula is given by

$F V = P {(1 + \frac{r}{n})}^{t \cdot n}$ .

Since we are compounding weekly you will use $n = 52$ . List out the other known values and then evaluate the formula for the amount of time the money will be in the account.

Answer (click to Show/Hide)

The only thing we need to change is the time, $t$ , in the compounding interest formula. We will start by listing out all the known values:

$\begin{array}{l} P = $ 200 \\ r = 0.07 \\ n = 52 \\ t = 4 \end{array}$

Next evaluate the compounding interest formula for $t = 4$ :

$\begin{array}{l} F V = 200 {(1 + \frac{0.07}{52})}^{(4 \cdot 52)} \\ F V \approx $ 264.58 \end{array}$

Now evaluate the formula for $t = 8$ :

$\begin{array}{l} F V = 200 {(1 + \frac{0.07}{52})}^{(8 \cdot 52)} \\ F V \approx $ 350.00 \end{array}$

One thing to note is that with the time doubling the amount of money earned interest is not doubling.

Rounding

It is important to be very careful about rounding when calculating expressions with exponents. In general, you want to keep as many decimals during calculations as you can if you are not able to evaluate the entire expression for a solution at once. Be sure to keep at least 3 significant digits (numbers after any leading zeros). Rounding 0.00012345 to 0.000123 will usually give you a “close enough” answer, but keeping more digits is always better.

To see why not over-rounding is so important, suppose you were investing $1000 at 5% interest compounded monthly for 30 years.

$\begin{array}{l} P = $ 1000 & the initial deposit \\ r = 0.05 & 5 % \\ n = 12 & 12 months in 1 year \\ t = 30 & since we ’ re looking for the amount after 3 0 years \end{array}$

If we first compute $\frac{r}{n}$ , we find $\frac{0.05}{12} = 0.00416666666667$ . Here is the effect of rounding this to different values:

$\frac{r}{n}$ rounded to:	Gives FV to be:	Error
0.004	$4208.59	$259.15
0.0042	$4521.45	$53.71
0.00417	$4473.09	$5.35
0.004167	$4468.28	$0.54
0.0041667	$4467.80	$0.06
no rounding	$4467.74

If you’re working in a bank, of course you wouldn’t round at all. For our purposes, the answer we got by rounding to 0.00417, three significant digits, is close enough – $5 off of $4500 isn’t too bad. Certainly keeping that fourth decimal place wouldn’t have hurt.

Applications with compounding discretely

Let us look at other applications with the compound interest formula. In some cases we may want to save up money for a goal. With the situation in the next example a fixed amount of money is deposited to reach a goal in the future. Remember this is a one time deposit that is being put into the account. We will look a situation where we make repeated deposits over time (like monthly deposits) in a later section.

Example 2

You know that you will need $40,000 for your child’s education in 18 years. If your account earns 4% compounded quarterly, how much would you need to deposit now to reach your goal?

Solution

In this example, we’re looking for an initial deposit $P$ . Start with listing out what is known:

$\begin{array}{l} r = 0.04 & 4 % \\ n = 4 & 4 quarters in 1 year \\ t = 18 & Since we know the balance in 18 years \\ FV = $ 40, 000 & The amount we have in 18 years \end{array}$

In this case, we’re going to have to set up the equation, and solve for $P$ .

$\begin{array}{l} 40000 = P {(1 + \frac{0.04}{4})}^{18 \cdot 4} \\ \frac{40000}{{(1 + \frac{0.04}{4})}^{18 \cdot 4}} = P \\ P ˜ $ 19, 539.84 \end{array}$

So you would need to deposit $19,539.84 now to have $40,000 in 18 years.

Try it Now 2

How much should be deposited today if the goal is to have $2000 five years from now when it is known that the account the money is to be deposited has an annual rate of 8.2% compounded daily?

Hint 1 (click to Show/Hide)

The question is looking for the initial deposit, $P$ . Use the compound interest formula to find $P$ when we now how much money is in the account when $t = 5$ .

Answer (click to Show/Hide)

Use the compounding interest formula to find $P$ , but first list out all the known values:

$\begin{array}{l} F V = $ 2000 \\ r = 0.082 \\ n = 365 \\ t = 5 \end{array}$

Next solve for $P$ :

$\begin{array}{l} F V = P {(1 + \frac{r}{n})}^{(t \cdot n)} \\ 2000 = P {(1 + \frac{0.082}{365})}^{(5 \cdot 365)} \\ P = \frac{2000}{{(1 + \frac{0.082}{365})}^{(5 \cdot 365)}} \\ P \approx $ 1, 327.36 \end{array}$

You would need to deposit $1,327.36 into the account to have $2000 in five years.

Compound Interest Continuously

Before we get into what we mean by compounding continuously lets look back at an example of compounding discretely. What we want to investigate is what happens as we compound more often during the same time frame.

Does compounding more frequently earn us more money?

The answer to this question is Yes, but there is some limiting factor. As we increase the number of times we compound we will earn more money, but there is an upper bound associated with the amount of money we can earn with the same interest rate compounded more often in the same time interval.

Let us look at what happens when we start with a principal of $3,200 in an account with an annual interest rate of 5.45% invested for 1.5 years and we vary the compounding period.

Table of values for different compounding periods for the same amount of time.
Compounding Rule	Number of times a year	Value in account after 1.5 years with an annual rate 5.45%
Yearly	$n = 1$	$P = 3200 {(1 + \frac{.0545}{1})}^{(1) (1.5)} = 3465.1326$
Quarterly	$n = 4$	$P = 3200 {(1 + \frac{.0545}{4})}^{(4) (1.5)} = 3470.6742$
Monthly	$n = 12$	$P = 3200 {(1 + \frac{.0545}{12})}^{(12) (1.5)} = 3471.9477$
Daily	$n = 365$	$P = 3200 {(1 + \frac{.0545}{365})}^{(365) (1.5)} = 3472.5691$
Hourly	$n = 8760$	$P = 3200 {(1 + \frac{.0545}{8760})}^{(8760) (1.5)} = 3472.5894$
Every Minute	$n = 525600$	$P = 3200 {(1 + \frac{.0545}{525600})}^{(525600) (1.5)} = 3472.5903$

Now what you should notice is that when we first started to increase the compounding period the gains in the increase in compounding were fairly noticeable, but look what happened when we went from monthly to daily. We are compounding by about 20 times more frequently, yet the gains are shrinking. This is even more noticeable as we go from hourly to every minute – this is compounding 60 times more frequently and the gain isn’t even measurable when we round to the nearest penny.

If we wish to compound every fraction of a second and then every fraction of that fraction of a second we are on our way to compounding continuously. The formula for continuous compounding is given below.

Compounding Interest Continuously

When interest is compounded continuously we can use the following formula to determine the future value of money:

$F V = P \cdot e^{(r \cdot t)}$

$F V$ is the future value of the account after $t$ years.
$P$ is the starting balance of the account (also called initial deposit, or principal).
$r$ is the annual interest rate in decimal form.
$t$ is the length of time the money sits in the account in years.
$e$ is a constant (like $π$ ) that is approximately 2.718 called Euler’s Number.

The most important thing to remember about using this formula is that it assumes that we put money in the account once and let it sit there earning interest while compounding continuously.

To finish up our example with depositing $3200 into an account with an annual rate of 5.45% for 1.5 years. We say that as we compounded more often in the year we approached some limiting value close to $3472.59. If we were to compound continuously we would have the following:

$\begin{array}{l} F V = P e^{r t} \\ F V = 3200 e^{(0.0545) (1.5)} \\ F V = 3200 (1.085184) \\ F V = 3472.590335 \\ F V ˜ $ 3472.59 \end{array}$

The actual value for compounding continuously compared to compounding every second was off by $0.00014 or less than a penny.

Evaluating exponents with e on the calculator

When we need to calculate something like $e^{3}$ it would be unwise to use an estimate like 2.718 to do the calculations if we can avoid it. As we increase the value on that exponent the estimation would become worse and worse.

Many scientific calculators have a button for $e$ or $e$ to an exponent. It is typically either labeled like $\begin{matrix} e \end{matrix}$ or $\begin{matrix} e^{x} \end{matrix}$

To evaluate $e^{3}$ we’d type $\begin{matrix} e \end{matrix}$ $\begin{matrix} ? \end{matrix}$ $3$ , or $\begin{matrix} e^{x} \end{matrix}$ $3$ .

Try it out – you should get something around 20.08553692.

If you don’t have the constant $e$ on your calculator use the approximation 2.718 for $e$ for now.

Example 3

Suppose you deposit $3000 in a CD paying 6% interest, compounded continuously. How much will you have in the account after 20 years? How did this compare to compounding Monthly?

Solution

In this scenario,

$\begin{array}{l} P & = & $ 3000 \\ \begin{array}{l} r \\ t \end{array} & \begin{array}{l} = \\ = \end{array} & \begin{array}{l} 0.06 \\ 20 \end{array} \end{array} \begin{array}{l} the initial deposit \\ \begin{array}{l} 6% annual rate \\ since we’re looking for how much we’ll have after 20 years \end{array} \end{array}$

So, $F V = 3000 e^{(0.06) (20)} ˜ $ 9, 960.35$ (round your answer to the nearest penny).

Compare this now to compounding monthly: $F V = 3000 {(1 + \frac{0.06}{12})}^{(20 \cdot 12)} ˜ $ 9, 930.61$

What we see is that by compounding continuously we have an additional gain of roughly $30 over that same time period (or roughly a 0.3% increase in value for compounding continuously).

Try it Now 3

If $3500 is invested at 9% compounded continuously, what will the value be in four years?

Hint 1 (click to Show/Hide)

The key word to look for in this problem is that the interest is being compounded continuously. This means we will be using $F V = P e^{r t}$ to calculate the amount in the account from an initial deposit. When using the formula make sure to always convert the percentage to a decimal before putting it into the formula.

Answer (click to Show/Hide)

Use the formula for compounding continuously with the given information:

$\begin{array}{l} F V = $ 3500 \\ t = 4 \\ r = .09 \\ F V = P e^{r t} \\ F V = 3500 e^{.09 \times 4} = $ 5016.65 \end{array}$

After four years there would be approximately $5016.65 in the account.

Like we did with compounding discretely we can solve for the initial amount needed to reach a goal (future value).

Example 4

You know that you will need $40,000 for your child’s education in 18 years. If your account earns 4% compounded continuously, how much would you need to deposit now to reach your goal?

Solution

In this example (as we saw in the last section), we’re looking for $P$ .

$\begin{array}{l} r & = & 0.04 & 4 % \\ t & = & 18 & Since we know the balance in 18 years \\ F V & = & $ 40, 000 & The amount we have in 18 years \end{array}$

Use the equation for continuous compounding to set up an equation to solve for P₀.

$\begin{array}{l} 40000 = P e^{0.04 \times 18} \\ 40000 = P (2.0544) \\ P = \frac{40000}{2.0544} = $ 19470.40 \end{array}$

You would need to deposit approximately $19,470.40 now to have $40,000 in 18 years.

As a side note you will have noticed we rounded $e^{0.04 \times 18}$ at the second step to 2.0544. This rounding introduced a small amount of error in our result when we go to divide 40,000 by 2.0544. If we had left the exact expression in and divided the total would have been $19.470.09 instead (see below):

$P = \frac{40000}{e^{0.04 \times 18}} = $ 19470.09$

Try it Now 4

The Tran family is expecing a new baby in a few days. They want to put away money for college and figure they will need about $100,000 in 18 years. If they make a single lump sum investment now into an account that compounds interest continuously with an annual rate of 3.4%, then how much would that be?

Hint 1 (click to Show/Hide)

In this question we are given the future value in the account and will be looking for the principal (the lump sum payment into the account at the start of the 18 years).

Answer (click to Show/Hide)

Find $P$ in the continuous compounding interest formula with the following from the question statement

$\begin{array}{l} F V = $ 100, 000 \\ r = 0.034 \\ t = 18 \end{array}$

Substitute the values into the formula $F V = P \cdot e^{(r \cdot t)}$ and solve for $P$ .

$\begin{array}{l} F V = P \cdot e^{(r \cdot t)} \\ 100, 000 = P e^{(0.034 \cdot 18)} \\ \frac{100, 000}{e^{(0.034 \cdot 18)}} = P \\ P ˜ $ 54, 226.53 \end{array}$

The Tran family needs to make a single lump sum deposit of $54,226.54 in the account.

Exercises

You deposit $300 in an account with an annual interest rate of 5% compounded annually. How much will you have in the account in 10 years?
Answer (click to Show/Hide)
This problem uses the formula for interest that is compounded a specific number of times per year.

Step 1: Identify the given values.
- Principal ( $P$ ): $300
- Interest Rate ( $r$ ): 5% or 0.05
- Compounding periods per year ( $n$ ): 1 (since it is compounded annually)
- Time ( $t$ ): 10 years
Step 2: Use the compound interest formula and solve.
The formula is $FV = P {(1 + \frac{r}{n})}^{n t}$ . We substitute our values into this formula.
$\begin{aligned} FV & = 300 {(1 + \frac{0.05}{1})}^{1 \times 10} \\ = 300 {(1.05)}^{10} \\ \approx 488.67 \end{aligned}$

Answer: You will have approximately $488.67 in the account in 10 years.
How much will $1000 deposited in an account with an annual interest rate of 7% compounded annually be worth in 20 years?
Answer (click to Show/Hide)
This problem uses the formula for interest that is compounded a specific number of times per year.

Step 1: Identify the given values.
- Principal ( $P$ ): $1000
- Interest Rate ( $r$ ): 7% or 0.07
- Compounding periods per year ( $n$ ): 1 (since it is compounded annually)
- Time ( $t$ ): 20 years
Step 2: Use the compound interest formula and solve.
The formula is $FV = P {(1 + \frac{r}{n})}^{n t}$ . We substitute our values into this formula.
$\begin{aligned} FV & = 1000 {(1 + \frac{0.07}{1})}^{1 \times 20} \\ = 1000 {(1.07)}^{20} \\ \approx 3869.68 \end{aligned}$

Answer: The account will be worth approximately $3,869.68 in 20 years.
You deposit $2000 in an account with an annual interest rate of 3% compounded monthly.
1. How much will you have in the account in 20 years?
2. How much interest will you earn?
Answer (click to Show/Hide)
This problem uses the formula for interest that is compounded a specific number of times per year.
1. How much will you have in the account in 20 years?
  Step 1: Identify the given values.
  - Principal ( $P$ ): $2000
  - Interest Rate ( $r$ ): 3% or 0.03
  - Compounding periods per year ( $n$ ): 12 (since it is compounded monthly)
  - Time ( $t$ ): 20 years
  Step 2: Use the compound interest formula and solve for Future Value (FV).
  The formula is $FV = P {(1 + \frac{r}{n})}^{n t}$ . We substitute our values into this formula.
  $\begin{aligned} FV & = 2000 {(1 + \frac{0.03}{12})}^{12 \times 20} \\ = 2000 {(1.0025)}^{240} \\ \approx 3641.51 \end{aligned}$
  Answer (a): The account will be worth approximately $3,641.51 in 20 years.
2. How much interest will you earn?
  Step 3: Calculate the interest earned.
  The interest earned is the difference between the future value and the initial principal.
  $Interest = FV - P$
  $Interest = 3641.51 - 2000 = 1641.51$
  
  Answer (b): You will earn approximately $1,641.51 in interest.
You deposit $10,000 in an account with an annual interest rate of 4% compounded monthly.
1. How much will you have in the account in 25 years?
2. How much interest will you earn?
Answer (click to Show/Hide)
This problem uses the formula for interest that is compounded a specific number of times per year.
1. How much will you have in the account in 25 years?
  Step 1: Identify the given values.
  - Principal ( $P$ ): $10,000
  - Interest Rate ( $r$ ): 4% or 0.04
  - Compounding periods per year ( $n$ ): 12 (since it is compounded monthly)
  - Time ( $t$ ): 25 years
  Step 2: Use the compound interest formula and solve for Future Value (FV).
  The formula is $FV = P {(1 + \frac{r}{n})}^{n t}$ . We substitute our values into this formula.
  $\begin{aligned} FV & = 10000 {(1 + \frac{0.04}{12})}^{12 \times 25} \\ = 10000 {(1.00333…)}^{300} \\ \approx 27137.65 \end{aligned}$
  Answer (a): The account will be worth approximately $27,137.65 in 25 years.
2. How much interest will you earn?
  Step 3: Calculate the interest earned.
  The interest earned is the difference between the future value and the initial principal.
  $Interest = FV - P$
  $Interest = 27137.65 - 10000 = 17137.65$
  
  Answer (b): You will earn approximately $17,137.65 in interest.
How much would you need to deposit in an account now in order to have $6,000 in the account in 8 years? Assume the account has an annual interest rate of 6% compounded monthly.
Answer (click to Show/Hide)
This problem requires us to solve for the Principal (P), which is the initial deposit, using the compound interest formula.

Step 1: Identify the given values.
- Future Value ( $FV$ ): $6,000
- Interest Rate ( $r$ ): 6% or 0.06
- Compounding periods per year ( $n$ ): 12 (since it is compounded monthly)
- Time ( $t$ ): 8 years
Step 2: Use the compound interest formula and solve for P.
The formula is $FV = P {(1 + \frac{r}{n})}^{n t}$ . We first substitute our values, then rearrange the formula to solve for P before calculating the final value.
$\begin{aligned} 6000 & = P {(1 + \frac{0.06}{12})}^{12 \times 8} \\ P & = \frac{6000}{{(1 + \frac{0.06}{12})}^{96}} \\ P & \approx 3717.14 \end{aligned}$

Answer: You would need to deposit approximately $3,717.14 in the account now.
How much would you need to deposit in an account now in order to have $20,000 in the account in 4 years? Assume the account has an annual interest rate of 5% interest compounded monthly.
Answer (click to Show/Hide)
This problem requires us to solve for the Principal (P), which is the initial deposit, using the compound interest formula.

Step 1: Identify the given values.
- Future Value ( $FV$ ): $20,000
- Interest Rate ( $r$ ): 5% or 0.05
- Compounding periods per year ( $n$ ): 12 (since it is compounded monthly)
- Time ( $t$ ): 4 years
Step 2: Use the compound interest formula and solve for P.
The formula is $FV = P {(1 + \frac{r}{n})}^{n t}$ . We first substitute our values, then rearrange the formula to solve for P before calculating the final value.
$\begin{aligned} 20000 & = P {(1 + \frac{0.05}{12})}^{12 \times 4} \\ P & = \frac{20000}{{(1 + \frac{0.05}{12})}^{48}} \\ P & \approx 16381.42 \end{aligned}$

Answer: You would need to deposit approximately $16,381.42 in the account now.
If $800 is invested in an account for 5 years compounded monthly with an annual rate of 4%, then would doubling the annual interest rate double the amount of interest earned? Explain.
Answer (click to Show/Hide)
To answer this question, we must calculate the interest earned in two separate scenarios: first with the original 4% rate, and second with a doubled rate of 8%. Then we can compare the results.

Scenario 1: Original Interest Rate (4%)

Step 1: Identify the given values.
- Principal ( $P$ ): $800
- Interest Rate ( $r$ ): 4% or 0.04
- Compounding periods per year ( $n$ ): 12 (monthly)
- Time ( $t$ ): 5 years
Step 2: Calculate the Future Value (FV).
$FV = 800 {(1 + \frac{0.04}{12})}^{12 \times 5} \approx 976.80$

Step 3: Calculate the interest earned.
$Interest = FV - P = 976.80 - 800 = 176.80$
The interest earned at 4% is $176.80.

Scenario 2: Doubled Interest Rate (8%)

Step 1: Identify the given values.
- Principal ( $P$ ): $800
- Interest Rate ( $r$ ): 8% or 0.08
- Other values ( $n$ and $t$ ) remain the same.
Step 2: Calculate the Future Value (FV).
$FV = 800 {(1 + \frac{0.08}{12})}^{12 \times 5} \approx 1191.88$

Step 3: Calculate the interest earned.
$Interest = FV - P = 1191.88 - 800 = 391.88$
The interest earned at 8% is $391.88.

Conclusion

No, doubling the annual interest rate does not double the amount of interest earned.
- Interest at 4%: $176.80
- Double the interest at 4%: $2 \times 176.80 = 353.60$
- Interest at 8%: $391.88
As we can see, the interest earned at 8% ($391.88) is more than double the interest earned at 4% ($353.60). This is due to the effect of compounding. With a higher interest rate, the “interest earned on interest” is also higher, which causes the investment to grow at a faster, exponential rate.
If $500 is invested in an account for 2 years compounded monthly with an annual rate of 3%, then would doubling the principal double the amount of interest earned? Explain.
Answer (click to Show/Hide)
To answer this question, we will calculate the interest earned in two scenarios: first with the original $500 principal, and second with a doubled principal of $1000. Then we will compare the results.

Scenario 1: Original Principal ($500)

Step 1: Identify the given values.
- Principal ( $P$ ): $500
- Interest Rate ( $r$ ): 3% or 0.03
- Compounding periods per year ( $n$ ): 12 (monthly)
- Time ( $t$ ): 2 years
Step 2: Calculate the Future Value (FV).
$FV = 500 {(1 + \frac{0.03}{12})}^{12 \times 2} \approx 530.88$

Step 3: Calculate the interest earned.
$Interest = FV - P = 530.88 - 500 = 30.88$
The interest earned with a $500 principal is $30.88.

Scenario 2: Doubled Principal ($1000)

Step 1: Identify the given values.
- Principal ( $P$ ): $1000
- Other values ( $r$ , $n$ , and $t$ ) remain the same.
Step 2: Calculate the Future Value (FV).
$FV = 1000 {(1 + \frac{0.03}{12})}^{12 \times 2} \approx 1061.76$

Step 3: Calculate the interest earned.
$Interest = FV - P = 1061.76 - 1000 = 61.76$
The interest earned with a $1000 principal is $61.76.

Conclusion

Yes, doubling the principal doubles the amount of interest earned.
- Interest with $500 principal: $30.88
- Double this interest: $2 \times 30.88 = 61.76$
- Interest with $1000 principal: $61.76
This occurs because the interest earned is directly proportional to the principal. The formula for interest can be written as $I = P [{(1 + \frac{r}{n})}^{n t} - 1]$ . Since the principal ( $P$ ) is a multiplier for the entire growth factor, doubling $P$ will also double the final interest amount.
Sam invests $1500 into a savings account in 2022. Ten years later anther $1500 is invested into the account. If we assume the interest is compounded monthly and has the same rate of the entire time of 4.5%, then how much is in the account 20 years after the original investment?
Answer (click to Show/Hide)
This problem involves two separate investments made at different times. We need to calculate the future value of each investment individually and then add them together to find the total amount in the account after 20 years.

Part 1: Calculate the Future Value of the First Investment

The first $1500 is invested for the full 20 years.

Step 1: Identify the given values.
- Principal ( $P$ ): $1500
- Interest Rate ( $r$ ): 4.5% or 0.045
- Compounding periods per year ( $n$ ): 12 (monthly)
- Time ( $t$ ): 20 years
Step 2: Calculate the Future Value (FV1).
$FV1 = 1500 {(1 + \frac{0.045}{12})}^{12 \times 20} \approx 3683.20$

Part 2: Calculate the Future Value of the Second Investment

The second $1500 is invested 10 years later, so it only earns interest for 10 years.

Step 1: Identify the given values.
- Principal ( $P$ ): $1500
- Interest Rate ( $r$ ): 4.5% or 0.045
- Compounding periods per year ( $n$ ): 12 (monthly)
- Time ( $t$ ): 10 years
Step 2: Calculate the Future Value (FV2).
$FV2 = 1500 {(1 + \frac{0.045}{12})}^{12 \times 10} \approx 2350.49$

Part 3: Find the Total Amount in the Account

To find the total value, we add the future values of both investments.
$Total FV = FV1 + FV2 = 3683.20 + 2350.49 = 6033.69$

Answer: After 20 years from the original investment, there will be approximately $6,033.69 in the account.
If $8,000 is invested at an annual rate of 9.2% compounded continuously, what will the final amount be in 4 years?
Answer (click to Show/Hide)
This problem requires the formula for continuously compounded interest.

Step 1: Identify the given values.
- Principal ( $P$ ): $8,000
- Interest Rate ( $r$ ): 9.2% or 0.092
- Time ( $t$ ): 4 years
Step 2: Use the continuous compounding formula.
The formula for the future value with interest compounded continuously is $FV = P e^{r t}$ , where $e$ is the mathematical constant approximately equal to 2.71828.

Step 3: Substitute the values into the formula and solve.
$\begin{aligned} FV & = 8000 e^{(0.092 \times 4)} \\ = 8000 e^{0.368} \\ \approx 11558.74 \end{aligned}$

Answer: The final amount after 4 years will be approximately $11,558.80.
What is the value of $750 invested into an account with an annual rate of 3.25% compounded continuously for 4 years and 3 months?
Answer (click to Show/Hide)
This problem requires the formula for continuously compounded interest.

Step 1: Identify the given values.
- Principal ( $P$ ): $750
- Interest Rate ( $r$ ): 3.25% or 0.0325
- Time ( $t$ ): 4 years and 3 months
Step 2: Convert the time to years.
The interest rate is an annual rate, so the time must also be in years.
$t = 4 years + 3 months \times (\frac{1 year}{12 months}) = 4 + 0.25 = 4.25 years$

Step 3: Use the continuous compounding formula.
The formula for the future value with interest compounded continuously is $FV = P e^{r t}$ .

Step 4: Substitute the values into the formula and solve.
$\begin{aligned} FV & = 750 e^{(0.0325 \times 4.25)} \\ = 750 e^{0.138125} \\ \approx 861.09 \end{aligned}$

Answer: The final amount after 4 years and 3 months will be approximately $861.09.
Find the formula to give the value of a deposit of $8,000 earning an annual rate of 9.2% compounded continuously for N years.
1. How much interest is earned after 2 years?
2. How much interest is earned after 4 years?
3. Were you surprised that it was not twice as much? Why or why not.
Answer (click to Show/Hide)
This problem uses the formula for continuously compounded interest, $FV = P e^{r t}$ .

Part 1: The Formula

First, we create the specific formula for this scenario by substituting the given values for Principal (P) and rate (r).
- Principal ( $P$ ): $8,000
- Interest Rate ( $r$ ): 9.2% or 0.092
- Time ( $t$ ): N years
The formula is:
$FV = 8000 e^{0.092 N}$
1. How much interest is earned after 2 years?
  Step 1: Calculate the Future Value (FV) for N = 2.
  $FV = 8000 e^{0.092 \times 2} = 8000 e^{0.184} \approx 9616.13$
  Step 2: Calculate the interest earned.
  Interest = FV – Principal = $9616.13 - 8000 = 1616.13$
  
  Answer (a): The interest earned after 2 years is approximately $1,616.13.
2. How much interest is earned after 4 years?
  Step 1: Calculate the Future Value (FV) for N = 4.
  $FV = 8000 e^{0.092 \times 4} = 8000 e^{0.368} \approx 11558.73$
  Step 2: Calculate the interest earned.
  Interest = FV – Principal = $11558.73 - 8000 = 3558.73$
  
  Answer (b): The interest earned after 4 years is approximately $3,558.73.
3. Were you surprised that it was not twice as much? Why or why not.
  It is not surprising that the interest earned after 4 years ($3,558.73) is more than double the interest earned after 2 years ( $2 \times 1616.13 = 3246.26$ ).
  
  This happens because of the nature of compounding interest. In the first two years, interest is earned only on the initial $8,000 principal. However, in the second two years (from year 2 to year 4), interest is earned on both the original principal and the accumulated interest from the first two years. This phenomenon of earning “interest on interest” causes the investment to grow at an accelerating rate, which is why the earnings in the second period are greater than in the first.
Sam’s grandparents was planning to give him $8,000 in 10 years. Sam has convinced his grandparents to pay him $4,000 now, instead. If Sam invests this $4,000 at an annual rate of 5.7% compounded continuously, how much money will he have in 10 years?
Answer (click to Show/Hide)
This problem requires the formula for continuously compounded interest to determine the future value of Sam’s investment.

Step 1: Identify the given values.
- Principal ( $P$ ): $4,000
- Interest Rate ( $r$ ): 5.7% or 0.057
- Time ( $t$ ): 10 years
Step 2: Use the continuous compounding formula.
The formula for the future value with interest compounded continuously is $FV = P e^{r t}$ .

Step 3: Substitute the values into the formula and solve.
$\begin{aligned} FV & = 4000 e^{(0.057 \times 10)} \\ = 4000 e^{0.57} \\ \approx 7073.07 \end{aligned}$

Answer: After 10 years, Sam will have approximately $7,073.07.
How much should be deposited into an account that has an annual rate of 4.15% compounded continuously so that in 15 years there is $10,000?
Answer (click to Show/Hide)
This problem requires the formula for continuously compounded interest, but we need to solve for the Principal (P), which is the initial deposit.

Step 1: Identify the given values.
- Future Value ( $FV$ ): $10,000
- Interest Rate ( $r$ ): 4.15% or 0.0415
- Time ( $t$ ): 15 years
Step 2: Use the continuous compounding formula and solve for P.
The formula is $FV = P e^{r t}$ . To find the principal, we can rearrange the formula algebraically:
$P = \frac{FV}{e^{r t}}$

Step 3: Substitute the values into the rearranged formula and solve.
$\begin{aligned} P & = \frac{10000}{e^{(0.0415 \times 15)}} \\ = \frac{10000}{e^{0.6225}} \\ \approx 5366.01 \end{aligned}$

Answer: You would need to deposit approximately $5,366.01 into the account now.
Rosa knows she needs $2350 for tuition in 9 months from now. How much should she deposit into an account that compounds continuously at an annual rate of 5.2% to achieve that goal?
Answer (click to Show/Hide)
This problem requires us to find the present value, or the initial deposit (P), needed to reach a specific future value with continuously compounding interest.

Step 1: Identify the given values.
- Future Value ( $FV$ ): $2350
- Interest Rate ( $r$ ): 5.2% or 0.052
- Time ( $t$ ): 9 months. We must convert this to years:
  $t = 9 months \times (\frac{1 year}{12 months}) = 0.75 years$
Step 2: Use the continuous compounding formula and solve for P.
The formula is $FV = P e^{r t}$ . To find the principal, we can rearrange the formula algebraically:
$P = \frac{FV}{e^{r t}}$

Step 3: Substitute the values into the rearranged formula and solve.
$\begin{aligned} P & = \frac{2350}{e^{(0.052 \times 0.75)}} \\ = \frac{2350}{e^{0.039}} \\ \approx 2260.11 \end{aligned}$

Answer: Rosa should deposit approximately $2,260.11 into the account.
Brandon plans on buying a car when he graduates in three years from now. He has some inheritance he can set away for this purchase. If he assumes the car will cost him $23,000 how much of that money should he set aside in an account that has an annual rate of 7.8% compounded continuously?
Answer (click to Show/Hide)
This problem requires us to find the present value, or the initial amount (P) Brandon should set aside to reach his goal with continuously compounding interest.

Step 1: Identify the given values.
- Future Value ( $FV$ ): $23,000
- Interest Rate ( $r$ ): 7.8% or 0.078
- Time ( $t$ ): 3 years
Step 2: Use the continuous compounding formula and solve for P.
The formula is $FV = P e^{r t}$ . To find the principal, we can rearrange the formula algebraically:
$P = \frac{FV}{e^{r t}}$

Step 3: Substitute the values into the rearranged formula and solve.
$\begin{aligned} P & = \frac{23000}{e^{(0.078 \times 3)}} \\ = \frac{23000}{e^{0.234}} \\ \approx 18201.32 \end{aligned}$

Answer: Brandon should set aside approximately $18,201.32 in the account.
As an experiment you plan on depositing $5 for your future great great grandchild to inherit 100 years from now. If the account compounds interest continuously with an annual rate of 5.5%, then how much will they inherit? What if you want to leave it for someone 200 years from now?
Answer (click to Show/Hide)
This problem requires the formula for continuously compounded interest, $FV = P e^{r t}$ , applied to two different time periods.

Step 1: Identify the given values.
- Principal ( $P$ ): $5
- Interest Rate ( $r$ ): 5.5% or 0.055
Part a: How much will they inherit in 100 years?

Step 2: Substitute the values into the formula with t = 100.
$\begin{aligned} FV & = 5 e^{(0.055 \times 100)} \\ = 5 e^{5.5} \\ \approx 1223.46 \end{aligned}$

Answer (a): In 100 years, your great great grandchild will inherit approximately $1,223.46.

Part b: How much will they inherit in 200 years?

Step 3: Substitute the values into the formula with t = 200.
$\begin{aligned} FV & = 5 e^{(0.055 \times 200)} \\ = 5 e^{11} \\ \approx 299370.71 \end{aligned}$

Answer (b): In 200 years, the inheritor would receive approximately $299,370.71.
Javier invests $500 into an account with the bank that compounds continuously with an annual rate of 9.0% and Maria invests $500 into a different account that compounds annually with an annual rate of 9.03%. Find the value in each account after 10 and 30 years. What did you notice? Should you compare more than just the annual rate to determine which account to invest in? Why or why not.
Answer (click to Show/Hide)
We need to calculate the future value for both Javier’s and Maria’s accounts at two different time intervals.

Part 1: Javier’s Account (Continuous Compounding)

We use the formula for continuously compounded interest: $FV = P e^{r t}$ .
- Principal ( $P$ ): $500
- Interest Rate ( $r$ ): 9.0% or 0.09
Value after 10 years:
$FV = 500 e^{(0.09 \times 10)} = 500 e^{0.9} \approx 1229.80$

Value after 30 years:
$FV = 500 e^{(0.09 \times 30)} = 500 e^{2.7} \approx 7439.87$

Part 2: Maria’s Account (Annual Compounding)

We use the formula for interest compounded n times per year: $FV = P {(1 + \frac{r}{n})}^{n t}$ .
- Principal ( $P$ ): $500
- Interest Rate ( $r$ ): 9.03% or 0.0903
- Compounding periods per year ( $n$ ): 1
Value after 10 years:
$FV = 500 {(1 + 0.0903)}^{10} \approx 1186.80$

Value after 30 years:
$FV = 500 {(1 + 0.0903)}^{30} \approx 6658.00$

Conclusion:

At both 10 years and 30 years, Javier’s account with continuous compounding has a higher value, even though it has a slightly lower annual interest rate. The difference becomes much more significant over the longer 30-year period.

Yes, you should always compare more than just the annual rate. The compounding frequency plays a crucial role in how quickly an investment grows. As this example shows, more frequent compounding (like continuous) can be more powerful than a slightly higher interest rate with less frequent compounding (like annual), especially over long investment horizons. This is because the interest earned begins to earn its own interest more often.

Attributions

This page contains modified content from David Lippman, “Math In Society, 2nd Edition.” Licensed under CC BY-SA 4.0.
This page contains content by Robert Foth, Math Faculty, Pima Community College, 2021. Licensed under CC BY 4.0.
Portions of the exercise solutions in this answer key were generated with the assistance of Gemini, a large language model from Google.

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Compound Interest Discretely

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Rounding

Applications with compounding discretely

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Compound Interest Continuously

Evaluating exponents with e on the calculator

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