Grade by Grade
C. Discrete Functions · Financial Applications: Interest, Present Value and Annuities

Finding the Interest Rate

C3.3About 65 minutes
Learning goals
  • I can solve A = P(1 + i)ⁿ for i using a rational exponent (an nth root).
  • I can convert the rate per compounding period into an annual rate.
  • I can find an average annual growth rate for a real-world amount that grew over time.

Words to know

rate per compounding period (i)
The interest rate for one period, as a decimal. For a stated annual rate compounded k times a year, i = annual rate ÷ k.
rational exponent
An exponent that is a fraction. x^(1/n) means the nth root of x.
average annual growth rate
The constant yearly percent rate that would turn the starting value into the final value over the same number of years.

What rate would do that?

Tyler's grandfather says, "When I was young I put $4 000 in the bank, and four years later I had $5 000." What interest rate was that? Here we know P, A and n. We need i, which is stuck inside a power. To free it we use a tool from the unit Exponential Functions: rational exponents.

Solving for i

Start with A = P(1 + i)ⁿ.

  1. Divide both sides by P: (1 + i)ⁿ = A ÷ P
  2. Raise both sides to the power 1/n (take the nth root): 1 + i = (A ÷ P)^(1/n)
  3. Subtract 1: i = (A ÷ P)^(1/n) − 1

This works because ((1 + i)ⁿ)^(1/n) = (1 + i)ⁿ × ¹ᐟⁿ = (1 + i)¹, by the power of a power law.

Remember: i is the rate per compounding period. If interest is compounded k times a year, the stated annual rate is k × i.

Undoing A = P(1 + i)ⁿ to find the rateA ÷ Pdivide out Pgives (1 + i)ⁿ^(1/n)undo the powergives 1 + i− 1subtract 1gives i× kannual rate
Each step undoes one operation in A = P(1 + i)ⁿ, working from the outside in. The last step changes the rate per period into an annual rate.
Worked example

Example 1: Annual compounding

$4 000 grew to $5 000 in 4 years, compounded annually. Find the annual interest rate.

  1. (1 + i)⁴ = 5000 ÷ 4000 = 1.25
  2. 1 + i = 1.25^(1/4)
  3. Calculator: 1.25 ^ (1 ÷ 4) = 1.057371…
  4. i ≈ 0.057371, so the rate is about 5.74% compounded annually.
  5. Check: 4000(1 + 0.057371)⁴ ≈ 5000 ✓

About 5.74% per year.

$4 000 growing at three rates (compounded annually)02 0004 0006 00001234567%5.74%: reaches $5 000 at year 44%target: $5 000time (years)amount ($)
Only the rate of about 5.74% passes through $5 000 at year 4. A lower rate falls short and a higher rate overshoots.
Worked example

Example 2: Monthly compounding

A savings account grew from $1 500 to $1 800 in 3 years, with interest compounded monthly. Find the annual interest rate.

  1. n = 3 × 12 = 36.
  2. (1 + i)³⁶ = 1800 ÷ 1500 = 1.2
  3. 1 + i = 1.2^(1/36) ≈ 1.0050773
  4. i ≈ 0.0050773 per month
  5. Annual rate = 12 × i ≈ 0.06093, about 6.09% compounded monthly.

About 6.09% compounded monthly.

Worked example

Example 3: Average annual growth rate

Suppose a hockey card bought for $40 in 2016 sold for $95 in 2026. Find its average annual growth rate.

  1. Treat the value as if it grew at a constant yearly rate (like compound interest compounded annually). n = 10 years.
  2. (1 + i)¹⁰ = 95 ÷ 40 = 2.375
  3. 1 + i = 2.375^(1/10) ≈ 1.09035
  4. i ≈ 9.04% per year.
  5. Note: the card did not really grow by exactly this percent each year. The average rate is the steady rate that gives the same start and end.

About 9.04% per year.

Worked example

Example 4: Quarterly compounding

A GIC turns $2 000 into $2 250 in 3 years, with interest compounded quarterly. What is the stated annual rate?

  1. n = 12 quarters. (1 + i)¹² = 2250 ÷ 2000 = 1.125
  2. i = 1.125^(1/12) − 1 ≈ 0.009864 per quarter
  3. Annual rate = 4i ≈ 3.95% compounded quarterly.

About 3.95% compounded quarterly.

Calculator check

Type the exponent in brackets: 1.2 ^ (1 ÷ 36). Without brackets, 1.2 ^ 1 ÷ 36 means (1.2¹) ÷ 36 = 0.0333, which is wrong. Keep at least 6 decimal places for i before multiplying by k, and give the final rate to 2 decimal places as a percent.

Just for fun

Riddle: what do you call a bank account that grew from $4 000 to $5 000 in four years?

A root-ine investment. (You need a fourth root to find its rate.)

Hands-on activity

What did it cost back then?

You need: a calculator, paper, and a grown-up to interview.

  1. Ask a grown-up to remember the price of something they bought when they were about your age (a movie ticket, a chocolate bar, a bus fare, a video game) and roughly what year it was.
  2. Find today's price for the same kind of item (a receipt, a store website, or a store visit).
  3. Find the number of years between the two prices and calculate the average annual growth rate: i = (new price ÷ old price)^(1/years) − 1.
  4. Compare with a classmate or another family member's item. Which price grew fastest?

What to look for (model answer): if a ticket cost $8 thirty years ago and costs $15 today, i = (15 ÷ 8)^(1/30) − 1 ≈ 0.0212, about 2.1% per year. Your answer depends on your item; a sensible answer is usually a few percent per year. Remember that memories of old prices are estimates.

Your turn

1

$1 000 grew to $1 500 in 8 years, compounded annually. Find the annual rate.

2

$2 400 grew to $3 000 in 5 years, compounded quarterly. Find the annual rate.

3

$5 000 grew to $6 200 in 5 years, compounded monthly. Find the annual rate.

4

$800 grew to $1 000 in 5 years, compounded annually. Find the rate.

5

What annual rate, compounded annually, doubles money in 9 years?

6

Rewrite 1 + i = ⁸√1.5 using a rational exponent.

More practice: warm-up

1

What does x^(1/4) mean?

2

Evaluate 1.21^(1/2).

3

If i = 0.005 per month, what is the stated annual rate?

4

If i = 0.0125 per quarter, what is the stated annual rate?

5

(1 + i)³ = 1.331. Find i.

More practice: on your own

1

A GIC turns $2 000 into $2 250 in 3 years, compounded quarterly. Find the stated annual rate.

2

Suppose a house was bought for $250 000 and sold 12 years later for $450 000. Find the average annual growth rate.

3

Suppose a litre of a product cost $3.50 fifteen years ago and $6.00 today. Find the average annual increase.

4

A student says the rate in Example 2 is 1.2^(1/36) − 1 ≈ 0.51%, so the answer is "0.51% per year". What is wrong?

5

Which grew faster: $1 000 to $1 300 in 4 years (compounded annually) or $1 000 to $1 600 in 8 years (compounded annually)?

Challenge

1

An investment triples in 15 years with interest compounded semi-annually. Find the stated annual rate.

2

Bank A offers 4.8% compounded monthly. What rate compounded annually would give exactly the same amount after one year? (This is the effective annual rate.)

3

$6 000 grew to $7 500 in 4 years with monthly compounding. Then the money was moved to a new account, and 2 years later it was worth $8 200, again with monthly compounding. Which account had the higher annual rate?

Check your understanding

Quiz

1.

Which expression gives i if A = P(1 + i)ⁿ?

2.

$500 grew to $605 in 2 years, compounded annually. What was the rate?

3.

A calculation gives i ≈ 0.0041 per month. The stated annual rate is about:

4.

Which calculator entry finds 1 + i when (1 + i)²⁴ = 1.3?

5.

What annual rate, compounded annually, doubles money in about 12 years?

For parents and teachers

Big idea: Expectation C3.3 asks students to solve A = P(1 + i)ⁿ for the interest rate per compounding period. This uses rational exponents from the unit Exponential Functions.

How to help: Have your teen explain each of the three "undo" steps (divide by P, take the nth root, subtract 1). Then ask, "Is that the rate per period or per year?"

Common difficulties:

  • Missing brackets around 1/n on the calculator.
  • Forgetting to subtract 1.
  • Reporting the monthly or quarterly rate as if it were the annual rate.
  • Rounding i too early; keep 6 or more decimal places until the end.