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6. Exponential & Logarithmic Functions
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United States of AmericaFL
Grade 912

6.07 Interest

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Practice
Lesson

Concept summary

Interest is generally calculated using one of two methods: simple interest or compound interest. It is typically calculated and/or paid at consistent time intervals, such as annually, monthly, daily, and so on.

Interest generated by simple interest accrues at a constant rate regardless of the time interval, so is a linear function. Interest generated by compound interest, however, accrues faster when the time interval is smaller, so is an exponential function.

Formulas for these methods of generating interest are as follows:

Simple interest

Interest calculated at any point in time using the principal value alone

\displaystyle A = P\left(1 + rt\right)
\bm{A}
The future value at the end of the time period
\bm{P}
The intial value or principal
\bm{r}
The interest rate
\bm{t}
The length of time

The unit of time for r and t must be the same.

Compound interest

Interest calculated using the cumulative value of the principal plus all interest accrued up to that point in time.

\displaystyle A = P\left(1 + \dfrac{r}{n}\right)^{nt}
\bm{A}
The future value at the end of the time period
\bm{P}
The intial value or principal
\bm{r}
The interest rate
\bm{n}
The number of compounding periods
\bm{t}
The length of time

If for compound interest the time interval (also called a compounding period) is shrunk down towards zero, we get the limiting case known as continuously compounded interest.

Continuously compounded interest

A = Pe^{rt}

\displaystyle A = Pe^{rt}
\bm{A}
The future value at the end of the time period
\bm{P}
The intial value or principal
\bm{e}
Euler's number, \approx 2.718 \ldots
\bm{r}
The interest rate
\bm{t}
The length of time

To compare compound interest to simple interest, we sometimes use a quantity called the annual percentage rate (or APR):

Annual percentage rate (APR)

A percentage simple interest rate which would generate an equivalent final value to a given compound interest scheme, on the same principal value over the same time period.

Worked examples

Example 1

Miles has \$11\,000 to invest in a savings account for 10 years, with a current interest rate of 4\% per year.

a

Determine the future value of the investment if interest is calculated using simple interest.

Approach

We have that:

  • P = 11\,000
  • r = 0.04
  • t = 10

and the simple interest formula is A = P\left(1 + rt\right).

Solution

\displaystyle A\displaystyle =\displaystyle P\left(1 + rt\right)State the formula
\displaystyle =\displaystyle 11\,000\left(1 + 0.04 \left(10\right) \right)Substitute known values
\displaystyle =\displaystyle 15\,400Evaluate the expression

So the investment will be worth a total of \$15\, 400 after 10 years if interest is calculated using simple interest.

b

Determine the future value of the investment if interest is calculated using compound interest, compounding quarterly.

Approach

We have that:

  • P = 11\,000
  • r = 0.04
  • t = 10
  • n = 4

and the compound interest formula is A = P\left(1 + \dfrac{r}{n}\right)^{nt}.

Solution

\displaystyle A\displaystyle =\displaystyle P\left(1 + \frac{r}{n}\right)^{nt}State the formula
\displaystyle =\displaystyle 11\,000\left(1 + \frac{0.04}{4}\right)^{4 \left(10\right)}Substitute known values
\displaystyle =\displaystyle 11\,000\left(1.01\right)^{40}Simplify the expression
\displaystyle =\displaystyle 16\,377.50Evaluate the expression

So the investment will be worth a total of \$16\,377.50 (to two decimal places) after 10 years if interest is calculated using compound interest, compounding quarterly.

c

Determine the future value of the investment if interest is calculated using continuously compounding interest.

Approach

We have that:

  • P = 11\,000
  • r = 0.04
  • t = 10

and the continously compounding interest formula is A = Pe^{rt}.

Solution

\displaystyle A\displaystyle =\displaystyle Pe^{rt}State the formula
\displaystyle =\displaystyle 11\,000 e^{0.04 \left(10\right)}Substitute known values
\displaystyle =\displaystyle 16\,410.07Evaluate the expression

So the investment will be worth a total of \$16\,410.07 (to two decimal places) after 10 years if interest is calculated using continuously compounding interest.

d

State which method of calculating interest generates Miles a larger amount of interest, and determine how much more this method earns than the other two methods.

Solution

We found that the final value after 10 years in each case is:

  • Simple interest: A = \$15\,400
  • Compound interest, compounding quarterly: A = \$16\,377.50
  • Continuously compounding interest: A = \$16\,410.07

So the method which generates the most interest is continuously compounding interest.

Over the same 10 year period, this method earned \$16\,410.07 - \$15\,400 = \$1\,010.07 more than using simple interest, and \$16\,410.07 - \$16\,377.50 = \$32.57 more than using compound interest.

Reflection

If the interest rates are equal, then compound interest will always earn more than simple interest over a fixed time period. In addition, a shorter compounding period will also result in more interest generated, and the limit of shrinking the compounding period is continuously compounding interest.

So since the same interest rate and time period was used in all three calculations, we expect that continuously compounding interest will have the largest result.

Example 2

An amount of \$760 is to be invested over a period of 6 years.

a

Determine the annual interest rate required to return \$960 from the investment if interest is being compounded monthly. Give your answer as a percentage to two decimal places.

Approach

We have the following information:

  • Final amount: A = 960
  • Principal: P = 760
  • Time period: t = 6
  • Number of compounding periods: n = 12
  • Interest rate per year: r = \,?

We can now substitute these values into the compound interest formula and rearrange to solve for the interest rate r.

Solution

\displaystyle A\displaystyle =\displaystyle P\left(1 + \frac{r}{n}\right)^{nt}State the formula
\displaystyle 960\displaystyle =\displaystyle 760\left(1 + \frac{r}{12}\right)^{12\left(6\right)}Substitute known values
\displaystyle \frac{960}{760}\displaystyle =\displaystyle \left(1 + \frac{r}{12}\right)^{72}Evaluate exponent and divide by 760
\displaystyle \left(\frac{960}{760}\right)^{\frac{1}{72}}\displaystyle =\displaystyle 1 + \frac{r}{12}Raise both sides to the power of \dfrac{1}{72}
\displaystyle \frac{r}{12}\displaystyle =\displaystyle \left(\frac{960}{760}\right)^{\frac{1}{72}} - 1Subtract 1 from both sides
\displaystyle r\displaystyle =\displaystyle 12\left(\frac{960}{760}\right)^{\frac{1}{72}} - 12Multiply both sides by 12
\displaystyle r\displaystyle =\displaystyle 0.0389990\ldotsEvaluate using a calculator

So the required annual interest rate as a percentage will be r = 3.90\%, rounded to two decimal places.

Reflection

Note that it is important for using this formula that the interest rate r and the time period t use the same unit for time.

In this case, the time period was 6 years, and we wanted to calculate the interest rate per year, so no conversion was needed.

b

Calculate the APR (annual percentage rate) of this investment. Give your answer as a percentage to two decimal places.

Approach

The APR for a compound interest scheme is the interest rate which would result in the same final value over the same time period if simple interest were used instead.

That is, given that:

  • Final amount: A = 960
  • Principal: P = 760
  • Time period: t = 6
  • Interest rate per year: r = \,?

we want to solve for the interest rate r using the simple interest formula.

Solution

\displaystyle A\displaystyle =\displaystyle P\left(1 + rt\right)State the formula
\displaystyle 960\displaystyle =\displaystyle 760\left(1 + 6r\right)Substitute known values
\displaystyle \frac{24}{19}\displaystyle =\displaystyle 1 + 6rDivide both sides by 760
\displaystyle 6r\displaystyle =\displaystyle \frac{5}{19}Subtract 1 from both sides
\displaystyle r\displaystyle =\displaystyle \frac{5}{114}Divide both sides by 6
\displaystyle r\displaystyle =\displaystyle 0.0438596\ldotsExpress as a decimal

So the equivalent APR as a percentage will be r = 4.39\%, rounded to two decimal places.

Example 3

The following table shows the value of an investment under two different conditions:

Investment 1Investment 2
Principal\$1200\$1200
Year 1\$1288.20\$1298.40
Year 2\$1382.88\$1396.80
Year 3\$1484.52\$1495.20
Year 4\$1593.64\$1593.60
Year 5\$1710.77\$1692.00
Year 6\$1836.51\$1790.40
a

Identify which investment is worth more after 6 years.

Solution

Looking at the "Year 6" row, we can see that Investment 1 is worth \$1836.51 while Investment 2 is worth \$1790.40. So the first investment is worth more after 6 years.

b

Identify how many years it takes for the investments to be worth approximately the same amount.

Solution

Comparing the value of each investment in each row (i.e. after every year), we can see that both investments are worth approximately \$1593.60 in the Year 4 row. So the two investments are worth the same amount after roughly 4 years.

c

Determine whether each investment is a form of simple or compound interest, explaining your reasoning.

Approach

An investment using simple interest will follow a linear pattern, while an investment using compound interest will follow an expoential pattern.

To identify which type of pattern is occuring for each investment, we can look at the differences between each pair of years.

Solution

The differences for Investment 1 are:

Investment 1Difference
Principal\$1200
Year 1\$1288.20\$88.20
Year 2\$1382.88\$94.68
Year 3\$1484.52\$101.64
Year 4\$1593.64\$109.12
Year 5\$1710.77\$117.13
Year 6\$1836.51\$125.74

The differences for Investment 2 are:

Investment 2Differences
Principal\$1200
Year 1\$1298.4098.40
Year 2\$1396.8098.40
Year 3\$1495.2098.40
Year 4\$1593.6098.40
Year 5\$1692.0098.40
Year 6\$1790.4098.40

We can see that the differences for Investment 1 are increasing at an increasing rate, which indicates that Investment 1 is using a form of compound interest.

The differences for Investment 2 are constant, so Investment 2 is using a form of simple interest.

Reflection

We can also confirm these patterns by plotting the values of each investment and fitting a curve to the data.

Investment 1
1
2
3
4
5
6
x
1100
1200
1300
1400
1500
1600
1700
1800
1900
y
Investment 2
1
2
3
4
5
6
x
1100
1200
1300
1400
1500
1600
1700
1800
1900
y

Outcomes

MA.912.AR.5.2

Solve one-variable equations involving logarithms or exponential expressions. Interpret solutions as viable in terms of the context and identify any extraneous solutions.

MA.912.AR.5.4

Write an exponential function to represent a relationship between two quantities from a graph, a written description or a table of values within a mathematical or real-world context.

MA.912.AR.5.7

Solve and graph mathematical and real-world problems that are modeled with exponential functions. Interpret key features and determine constraints in terms of the context.

MA.912.AR.5.9

Solve and graph mathematical and real-world problems that are modeled with logarithmic functions. Interpret key features and determine constraints in terms of the context.

MA.912.FL.3.1

Compare simple, compound and continuously compounded interest over time.

MA.912.FL.3.2

Solve real-world problems involving simple, compound and continuously compounded interest.

MA.912.FL.3.4

Explain the relationship between simple interest and linear growth. Explain the relationship between compound interest and exponential growth and the relationship between continuously compounded interest and exponential growth.

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