Applications: Compound Interest, Half-Life, Population

By the end of this lesson, you will be able to apply exponential growth and decay formulas to compound interest, half-life, and population models; solve for any variable in these models; and interpret the results in context.

The Big Idea

Exponential functions are the mathematical language of processes that change proportionally to their current size. Three of the most important applications in finance, physics, and biology use nearly identical mathematics: only the parameters and context differ. Mastering the three formulas and knowing when to apply each is the goal.

Formula 1: Compound Interest

Compounding $n$ times per year:

$$A = P\left(1 + \frac{r}{n}\right)^{nt}$$

Continuous compounding:

$$A = Pe^{rt}$$

• $P$ = principal (initial investment)
• $r$ = annual interest rate (as a decimal)
• $n$ = compounding periods per year
• $t$ = time in years
• $A$ = amount after $t$ years

Formula 2: Exponential Growth/Decay

$$A = A_0 e^{kt}$$

• $A_0$ = initial amount
• $k > 0$: growth; $k < 0$: decay
• $t$ = time
• $A$ = amount at time $t$

Formula 3: Half-Life

$$A = A_0 \left(\frac{1}{2}\right)^{t/h}$$

• $h$ = half-life (time for quantity to halve)
• $t$ = elapsed time

Equivalently: $k = -\dfrac{\ln 2}{h}$ in the exponential form.

Example 1 — Compound Interest

$5,000 invested at 6% compounded monthly for 10 years. Find the final amount.

$$A = 5000\left(1 + \frac{0.06}{12}\right)^{120} = 5000(1.005)^{120} \approx 5000(1.8194) \approx $9,097$$

Example 2 — Continuous Compounding

$3,000 at 4% continuously for 15 years.

$$A = 3000e^{0.04 \times 15} = 3000e^{0.6} \approx 3000(1.8221) \approx $5,466$$

Example 3 — Doubling Time

At what interest rate (compounded continuously) does money double in 12 years?

$2P = Pe^{12r} \implies 2 = e^{12r} \implies \ln 2 = 12r \implies r = \dfrac{\ln 2}{12} \approx 0.0578 = 5.78%$

Rule of 72: The doubling time at rate $r%$ is approximately $\dfrac{72}{r}$ years. At 6%: $72/6 = 12$ years — close to our exact answer.