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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
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$1,60,000$
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
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
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