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Exponential growth and decay

Model repeated percentage change with multipliers, compound interest and exponential functions.

Also known as: compound interest, depreciation, exponential functions.

Key ideas

  • Growth by r%r\% uses the multiplier 1+r1001 + \frac{r}{100}; decay uses 1−r1001 - \frac{r}{100}.
  • After nn periods: value=start×multipliern\text{value} = \text{start} \times \text{multiplier}^n.
  • f(t)=a⋅btf(t) = a \cdot b^t is growth when b>1b > 1 and decay when 0<b<10 < b < 1.
  • Exponential models change by a constant factor; linear models change by a constant amount.

Multipliers

Increasing by 15%15\% is the same as multiplying by 1.151.15. Decreasing by 15%15\% means multiplying by 0.850.85.

Repeated change uses powers. A value of 12001200 that falls by 15%15\% each year is worth V=1200×0.85tV = 1200 \times 0.85^t after tt years.

Compound interest

Compound interest adds interest to the balance, so each year's interest is larger. At 3%3\% per year on 20002000: 2000×1.032=2121.802000 \times 1.03^2 = 2121.80 after two years, compared with 21202120 using simple interest.

Reading exponential functions

In f(t)=a⋅bt/kf(t) = a \cdot b^{t/k}, aa is the starting value and the quantity is multiplied by bb every kk units of time. For f(t)=80(1.5)t/2f(t) = 80(1.5)^{t/2}, the value is multiplied by 1.51.5 every 22 units, so by 1.52=2.251.5^2 = 2.25 every 44 units.

Worked examples

Example 1 · Depreciation

A car bought for 15 000 loses 20% of its value each year. What is it worth after 3 years?

  1. Step 1.Multiplier for a 20%20\% decrease: 0.80.8.
  2. Step 2.15000×0.83=15000×0.51215000 \times 0.8^3 = 15000 \times 0.512.

Answer: 7680

Example 2 · Doubling

A population of 500 doubles every year. What is it after 3 years?

  1. Step 1.500×23=500×8500 \times 2^3 = 500 \times 8.

Answer: 4000

Common mistakes

  • Using 0.150.15 instead of 0.850.85 for a 15%15\% decrease.
  • Applying simple interest (adding the same amount each year) to a compound problem.
  • Treating 1200(0.85)t1200(0.85)^t as a linear decrease of 0.150.15 per year.

Try it yourself

Original practice questions. Your answer is checked instantly.

Question 1Exponential growth and decayEasier

A population of 500 bacteria doubles every hour. How many bacteria are there after 3 hours?

Question 2Exponential growth and decayEasier

Which of the following functions shows exponential growth?

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