Question

A population grows at the rate of 5% per year. How long does it take for the population to double?

Answer 1

Let P₀ be the initial population and P be the population at any time t. Then,
$$\begin{gathered} \frac{dP}{dt}=\frac{5P}{100} \\ \Rightarrow \frac{dP}{dt}=0.05P \end{gathered}$$

$$\begin{gathered} \Rightarrow \frac{dP}{P}=0.05dt \\ \mathrm{Integrating}\mathrm{both}\mathrm{sides}\mathrm{with}\mathrm{respect}\mathrm{to}t,\mathrm{we}\mathrm{get} \\ \int \frac{dP}{P}=\int 0.05dt \\ \log P=0.05t+C \\ \mathrm{Now}, \\ P=P_0\mathrm{at}t=0 \\ \therefore \log P_0=0+C \\ \Rightarrow C=\log P_0 \\ \mathrm{Putting}\mathrm{the}\mathrm{value}\mathrm{of}C,\mathrm{we}\mathrm{get} \\ \log P=0.05t+\log P_0 \\ \Rightarrow \log \frac{P}{P_0}=0.05t \\ \mathrm{To}\mathrm{find}\mathrm{the}\mathrm{time}\mathrm{when}\mathrm{the}\mathrm{population}\mathrm{will}\mathrm{double},\mathrm{we}\mathrm{have} \\ P=2P_0 \\ \therefore \log \frac{2P_0}{P_0}=0.05t \\ \Rightarrow \log 2=0.05t \\ \Rightarrow t=\frac{\log 2}{0.05}=20\log 2\mathrm{years} \end{gathered}$$