The population of a village increases at a rate proportional to the population at that time. In a period of 10 years the population grew from 20,000 to 40,000 , then the population after another 20 years is

We have \(\frac{\mathrm{dP}}{\mathrm{dt}} \propto \mathrm{P} \Rightarrow \frac{\mathrm{dP}}{\mathrm{dt}}=\mathrm{kP} \quad \therefore \int \frac{\mathrm{dP}}{\mathrm{P}}=\int \mathrm{kdt} \Rightarrow \log \mathrm{P}=\mathrm{kt}+\log \mathrm{c}\)
We have: when \(t=0, P=20000\) and when \(t=10, P=40000\)
\(\therefore \quad \log 20000=\log \mathrm{c}\)
Now \(\log 40000=10 \mathrm{k}+\log 20000\)
\(\therefore \log \left(\frac{40000}{20000}\right)=10 \mathrm{k} \Rightarrow \mathrm{k}=\frac{1}{10} \log 2\)
\(\therefore \log P=\left(\frac{1}{10} \log 2\right) t+\log 20000\)
When \(t=30\), we write
\(\log P=\frac{30}{10} \log 2+\log 20000=\log [(8)(20000)]\) \(\therefore P=160000\)
This problem can also be solved as follows :
From given data, we say that population doubles in 10 years.
\(\therefore\) In next 10 years, population changes from 40,000 to 80,000 .
In 10 years after that, population changes from 80000 to 160000