Percent changes that repeat are exponential. You know percent increase and decrease; in an exponential function that percent becomes the base.
Why it works
If a population grows $5\%$ in a year, the new population is the old one plus $5\%$ of it: $P + 0.05P = P(1 + 0.05) = 1.05P$. So one year of growth means multiplying by $1.05$, and $t$ years means multiplying by $1.05$ again and again: $P_0 \cdot 1.05^t$. A $5\%$ decrease keeps $95\%$: multiply by $1 - 0.05 = 0.95$. Each year's change is a percent of the current amount, which is why the yearly change grows (or shrinks) over time.
| Change per period | Growth or decay factor $b$ |
|---|---|
| grows $5\%$ | $1 + 0.05 = 1.05$ |
| grows $20\%$ | $1.2$ |
| shrinks $5\%$ | $1 - 0.05 = 0.95$ |
| shrinks $30\%$ | $0.7$ |
| doubles | $2$ |
| halves | $0.5$ |
Key idea
A growth rate $r$ gives the factor $b = 1 + r$; a decay rate $r$ gives $b = 1 - r$. The model is $A = A_0 \cdot b^t$.
Worked examples
Worked example: Growth
A painting worth \$1000 gains $10\%$ of its value each year. What is it worth after $3$ years?
- Factor: $1 + 0.10 = 1.1$.
- $1000 \cdot 1.1^3 = 1000 \cdot 1.331 = 1331$.
Answer: \$1331 (not \$1300: each year's $10\%$ is of a larger amount).
Worked example: Decay
A car worth \$20,000 loses $15\%$ of its value per year. Write a model and find the rate from a model.
- Factor: $1 - 0.15 = 0.85$, so $V = 20000 \cdot 0.85^t$.
- Going the other way, $A = 50 \cdot 0.92^t$ keeps $92\%$ each year: a decay rate of $8\%$.
Answer: $V = 20000 \cdot 0.85^t$; $0.92$ means $8\%$ decay.
Common mistake
Using $0.05$ as the growth factor for $5\%$ growth, or $1.05$ for a $5\%$ decrease. Multiplying by $0.05$ would leave only $5\%$ of the amount. Check: after one period the amount should change only a little, so the factor should be close to $1$, above it for growth and below it for decay.