B. Exponential Functions · Financial Mathematics: Compound Interest and AnnuitiesCompound Interest Is Exponential Growth
B3.3 B3.1About 70 minutes
Learning goals- I can show with a spreadsheet, a table and a graph that compound interest grows exponentially.
- I can match A = P(1 + i)ⁿ to the exponential function f(n) = a(b)ⁿ and say what a and b mean.
- I can describe the domain and range of a compound interest function in context.
- I can use the constant doubling time of exponential growth, and estimate it with the Rule of 72.
Words to know
exponential function
A function of the form f(x) = a(b)ˣ with a ≠ 0, b > 0 and b ≠ 1. The variable is in the exponent.
growth factor (base, b)
The number the value is multiplied by each period. For compound interest, b = 1 + i.
initial value (a)
The value when the exponent is 0. For compound interest, a = P.
discrete
Made of separate values. Compound interest is added only at the end of each period, so n takes whole-number values.
doubling time
The time it takes an amount to double. For exponential growth it is the same no matter where you start.
Rule of 72
A quick estimate: doubling time in years ≈ 72 ÷ (annual rate as a percent), for interest compounded annually.
Same shape, different stories
A population of bacteria, the value of a rare hockey card, and the balance in a savings account can all follow the same kind of curve. In the unit Exponential Functions you studied functions such as f(x) = 2ˣ and their key properties. In this lesson you will see that every compound interest problem is an exponential function in disguise, and that the properties you learned there tell you real things about money.
Matching the formulas
Compare f(n) = a(b)ⁿ with A(n) = P(1 + i)ⁿ:
- the initial value a is the principal P (the value when n = 0);
- the base b is the growth factor 1 + i, which is greater than 1, so the function increases;
- the exponent is n, the number of compounding periods.
For example, $500 at 8% compounded annually is A(n) = 500(1.08)ⁿ: initial value 500, growth factor 1.08.
The present value formula is also exponential: PV(n) = A(1 + i)⁻ⁿ = A(1/(1 + i))ⁿ, with a base between 0 and 1, so it decreases (exponential decay).
How to tell linear from exponential in a table
- Constant first differences (the same amount added each step) → linear. This is simple interest.
- Constant ratios (the same factor multiplied each step) → exponential. This is compound interest.
The first differences of an exponential table are not constant, and neither are the second differences, but the ratios are.
Worked exampleExample 1: Reading an exponential function
An account balance is modelled by A(n) = 2500(1.0125)ⁿ, where interest is compounded quarterly and n is the number of quarters. (a) What was deposited? (b) What is the interest rate per quarter, and per year? (c) What is the balance after 2 years?
- (a) The initial value is a = 2500, so $2 500 was deposited.
- (b) The base is 1 + i = 1.0125, so i = 0.0125 = 1.25% per quarter. Per year: 4 × 1.25% = 5% compounded quarterly.
- (c) 2 years = 8 quarters. A(8) = 2500(1.0125)⁸ ≈ 2500 × 1.1044861 ≈ $2 761.22.
$2 500 deposited at 5% per year compounded quarterly; about $2 761.22 after 2 years.
Worked exampleExample 2: Is this compound interest? Find the equation.
An investment's values at the end of years 0 to 4 are: $250.00, $265.00, $280.90, $297.75, $315.62. Decide whether it grows exponentially, and write an equation.
- First differences: 15.00, 15.90, 16.85, 17.87. Not constant, so not linear.
- Ratios: 265 ÷ 250 = 1.06; 280.90 ÷ 265 = 1.06; 297.75 ÷ 280.90 ≈ 1.06; 315.62 ÷ 297.75 ≈ 1.06. Constant ratio, so exponential.
- Initial value a = 250; growth factor b = 1.06, so i = 0.06.
Exponential growth: A(n) = 250(1.06)ⁿ, which is $250 invested at 6% compounded annually.
All three graphs start at the same initial value, $1 000. A bigger growth factor (base) makes the curve rise much faster: after 30 years, 8% gives more than twice what 5% gives, and more than five times what 2% gives.Domain and range in context
The function f(x) = 1000(1.05)ˣ from the unit Exponential Functions has domain {x ∈ ℝ} and range {y ∈ ℝ | y > 0}, with a horizontal asymptote y = 0. In a money problem the context restricts these:
- Interest is only added at the end of each period, so n is a whole number: n = 0, 1, 2, … up to the end of the term. The graph is really a set of points. We often join them with a smooth curve to see the pattern, but between compounding dates the balance does not change.
- The range is only the values the account actually takes, from P up to the final amount.
Example: $1 200 at 4.5% compounded annually for 10 years. A(n) = 1200(1.045)ⁿ. Domain: {0, 1, 2, …, 10}. Range: {1200, 1254, …, 1863.56} (the 11 values of A(n)), each between $1 200 and $1 863.56.
Doubling time
A special property of exponential growth: the time it takes to double is always the same, whatever amount you start with. $500 at 8% compounded annually is about $1 000 after 9 years, about $2 000 after 18 years, and about $4 000 after 27 years.
You met the Rule of 72 in Grade 8: for annual compounding, doubling time ≈ 72 ÷ (rate as a percent). At 8%: 72 ÷ 8 = 9 years. Check: 1.08⁹ ≈ 1.999. It is only an estimate, but a very good one for rates between about 4% and 12%.
Worked exampleExample 3: Using doubling time
Use the Rule of 72 to estimate how long $3 000 takes to grow to $12 000 at 6% compounded annually. Then check with the formula.
- $3 000 → $12 000 is two doublings (× 2 × 2 = × 4).
- Doubling time ≈ 72 ÷ 6 = 12 years, so two doublings take about 24 years.
- Check: 3000(1.06)²⁴ ≈ 3000 × 4.0489 ≈ 12 146.80. Just over $12 000. ✓
About 24 years. The formula confirms that after 24 years the amount is about $12 146.80.
Just for funThere is an old legend about a chessboard: put 1 grain of rice on the first square, 2 on the second, 4 on the third, and keep doubling. The 64th square alone would need 2⁶³ grains, about 9.2 quintillion. That is more than a century's worth of the whole world's rice harvest, on a single square. Exponential growth is not shy.
Hands-on activitySpreadsheet investigation
You need: a computer or tablet with a free spreadsheet (Google Sheets, Excel or LibreOffice Calc).
- In A1 type
n, in B1 type Amount, in C1 Ratio, in D1 First difference. - Type 0 to 20 in column A (A2 to A22).
- In B2 type
500. In B3 type =B2*1.08 and fill down to B22. - In C3 type
=B3/B2 and in D3 =B3-B2; fill both down. - Make a scatter chart of columns A and B.
- Change the 1.08 to 1.03 and to 1.12 (edit B3 and fill down again). Watch the chart.
- Change B2 from 500 to 1000. What happens to every value?
What to look for: column C is always the growth factor; column D keeps growing. A bigger growth factor makes the graph curve upward faster. Doubling the principal doubles every amount (a vertical stretch by a factor of 2), but it does not change the doubling time.
Your turn
1A(n) = 1800(1.04)ⁿ, compounded annually. State the principal, the annual rate, and A(5).
Principal $1 800, rate 4% per year. A(5) = 1800(1.04)⁵ ≈ $2 189.98.
2A(n) = 600(1.005)ⁿ where n is in months. What is the annual rate, and how often is interest compounded?
i = 0.005 per month, so 6% per year compounded monthly.
3Values: $800, $824, $848.72, $874.18. Linear or exponential? Write the equation.
Ratios are all 1.03, so exponential: A(n) = 800(1.03)ⁿ.
4Values: $800, $824, $848, $872. Linear or exponential? What kind of interest?
First differences are all $24: linear, so simple interest at 3% (24 ÷ 800).
5Use the Rule of 72 to estimate the doubling time at 9% compounded annually. Check with 1.09⁸.
72 ÷ 9 = 8 years. 1.09⁸ ≈ 1.993, so after 8 years the amount has almost exactly doubled. ✓
6State the domain and range of A(n) = 2000(1.05)ⁿ for a 4-year investment compounded annually.
Domain: {0, 1, 2, 3, 4}. Range: {2000, 2100, 2205, 2315.25, 2431.01} (to the cent).
More practice: warm-up
1In A = P(1 + i)ⁿ, which letter plays the role of the base b in f(n) = a(b)ⁿ?
2What is the growth factor for 7% compounded annually?
3Does PV = 1000(1.04)⁻ⁿ show growth or decay as n increases?
Decay: the base 1/1.04 is less than 1.
4Rule of 72: estimate the doubling time at 4%.
5Why do we say n is discrete for compound interest?
Interest is only added at the end of each period, so n takes whole-number values.
More practice: on your own
1$3 000 is invested at 6% compounded quarterly. Write A(n) and find A after 10 years.
A(n) = 3000(1.015)ⁿ, n in quarters. 10 years = 40 quarters: A(40) ≈ $5 442.06.
2Two accounts start with $1 000. One grows by 1.04 each year, the other by 1.06. Which reaches $2 000 first, and roughly when?
Rule of 72: 72 ÷ 6 = 12 years for the 1.06 account; 72 ÷ 4 = 18 years for the 1.04 account. The 6% account reaches $2 000 first, after about 12 years (1.06¹² ≈ 2.012).
3A $5 000 investment doubles every 12 years. Without a formula, how much is it worth after 36 years?
36 years is 3 doublings: 5000 × 2 × 2 × 2 = $40 000.
4Explain why changing the principal from $500 to $1 500 does not change the doubling time.
Doubling time depends only on the growth factor: you need (1 + i)ⁿ = 2. The principal cancels out: P(1 + i)ⁿ = 2P gives (1 + i)ⁿ = 2. Tripling P triples every value but leaves the doubling time the same.
5Write the present value of $8 000 as a function of n at 5% compounded annually. Find PV(0), PV(5) and PV(10).
PV(n) = 8000(1.05)⁻ⁿ. PV(0) = $8 000.00, PV(5) ≈ $6 268.21, PV(10) ≈ $4 911.31.
Challenge
1An account grows from $2 000 to $2 676.45 in 5 years, compounded annually. Show that this fits A(n) = 2000(b)ⁿ and find b and the annual rate. (Hint: b⁵ = 2676.45 ÷ 2000. Use rational exponents from the unit Exponential Functions.)
b⁵ = 2676.45 ÷ 2000 = 1.338225. So b = 1.338225^(1/5) ≈ 1.06 and the rate is 6% per year. Check: 2000(1.06)⁵ ≈ $2 676.45 ✓. You will use this method again in the lesson on finding the interest rate.
2A $1 000 investment at 8% compounded annually and another $1 000 investment at 4% compounded annually are both held for 36 years. Use doubling times to compare them without a calculator.
At 8%: doubling time ≈ 9 years, so 4 doublings in 36 years: about $16 000. At 4%: doubling time ≈ 18 years, so 2 doublings: about $4 000. Doubling the rate made the result about 4 times as big, not twice as big.
3The graph of A(n) = P(1 + i)ⁿ has a horizontal asymptote at A = 0 as n → −∞. What would a negative n mean in a money context? Does it make sense?
A negative n means going back in time: A(−3) = P(1 + i)⁻³ is the amount that, invested 3 periods ago, grows to P now. That is a present value calculation, so it can make sense, but in a problem that starts today, the domain is n ≥ 0.
Quiz
1. In A(n) = 750(1.035)ⁿ, compounded annually, what is the annual interest rate?
2. Which table shows exponential growth?
3. For an investment compounded annually for 6 years, the domain of A(n) is:
4. Using the Rule of 72, money at 12% compounded annually doubles in about:
5. $400 doubles every 10 years. What is it worth after 30 years?
6. Which is true about the first differences of a compound interest table?
Big idea (B3.3): Students determine, through investigation with spreadsheets and graphs, that compound interest is an example of exponential growth: A = P(1 + i)ⁿ and PV = A(1 + i)⁻ⁿ are exponential functions of n. This connects this unit to the unit Exponential Functions. B3.1 (tables and graphs comparing simple and compound interest) is revisited through first differences and ratios.
How to help: If possible, let your teen build the spreadsheet in the activity and show you what happens when the rate changes. Ask: "How can you tell from a table whether it is simple or compound interest?" (Constant differences versus constant ratios.)
Common difficulties:
- Confusing the rate i with the growth factor 1 + i.
- Treating the Rule of 72 as exact; it is an estimate for annual compounding.
- Forgetting that in context n is a whole number, so the graph is really points.
In Grade 12 (Advanced Functions, MHF4U) students learn logarithms, which solve exponential equations such as (1.08)ⁿ = 2 exactly.