Exponential functions model things that change by a fixed percent: radioactive decay, medicine leaving the body, money and populations. Solving with logs answers the natural question, "how long until…?"
Key idea
Half-life $h$: $A = A_0\left(\frac{1}{2}\right)^{t/h}$. Continuous growth or decay: $A = A_0e^{kt}$ (with $k > 0$ for growth, $k < 0$ for decay). Doubling time for continuous rate $r$: $t = \frac{\ln 2}{r}$. Half-life for $A_0e^{-kt}$: $h = \frac{\ln 2}{k}$.
Why it works
"Half-life $h$" means every $h$ units of time multiply the amount by $\frac{1}{2}$. In $t$ units there are $\frac{t}{h}$ such halvings, which gives the first formula. For doubling, set $A = 2A_0$ in $A_0e^{rt}$: $e^{rt} = 2$, so $rt = \ln 2$. Since $\ln 2 \approx 0.693$, a 7% rate doubles in about $\frac{0.693}{0.07} \approx 10$ years: the "rule of 70."
Worked examples
Worked example: Whole half-lives
A sample of 200 g has a half-life of 8 years. How much is left after 24 years?
- $\frac{24}{8} = 3$ half-lives.
- $200\left(\frac{1}{2}\right)^3 = 25$.
Answer: 25 g
Worked example: How long until 30% remains?
Same substance. When will 30% remain?
- $0.3 = \left(\frac{1}{2}\right)^{t/8}$. Take logs: $\ln 0.3 = \frac{t}{8}\ln 0.5$.
- $t = 8 \cdot \frac{\ln 0.3}{\ln 0.5} \approx 13.90$.
Answer: about 13.9 years
Worked example: Doubling time
A population grows continuously at 3% per year. When does it double?
- $e^{0.03t} = 2$, so $t = \frac{\ln 2}{0.03} \approx 23.1$.
Answer: about 23.1 years
A quick check for any model: plug in $t = 0$ (you should get the starting amount) and one half-life or doubling time (you should get half or double). If those two checks pass, the formula is set up correctly.
Common mistake
Treating decay as subtraction: "half-life 8 years, so after 16 years it is all gone." Each half-life halves what is left, so after 16 years a quarter remains. Exponential decay never quite reaches zero.