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The population of a country doubles in 40 years. Assuming that the rate of increase is proportional to number of inhabitants, find the number of years in which the population would treble itself?

$\begin{array}{1 1}(A) \;20 \large\frac{\log 2}{\log 3}\; years \\(B) \;40 \large\frac{\log 2}{\log 3}\; years \\ (C) \;40 \large\frac{\log 3}{\log 2}\; years \\ (D) \;20 \large\frac{\log 3}{\log 2}\; years\end{array}$

1 Answer

Let the original population by $P$. Let the population at the end of $t$ years by $x$.
Then, $\large\frac{dx}{dt} $$\propto x = kx$, where $k$ is a constant
Integrating, we get, $\log x = kt + C$
Initially, at $t = 0$, $x = P \rightarrow \log P = C \rightarrow \log x = kt + \log P$ .. (1)
At $t = 40$, $x = 2P$, $\log 2P = 40k + \log P \rightarrow 40k = \log 2 \rightarrow k = \large\frac{\log 2}{40}$
From (1), we get, $\log x= \large \frac{1}{40} $$\log 2\times t + \log P$
If $x = 3P \rightarrow \log 3P = \large\frac{1}{40}$$ \log 2 \times t + \log P$
$\Rightarrow \large\frac{1}{40}$$ \log 2 \times t = \log 3P - \log P = \log 3$
$\Rightarrow t = 40 \large\frac{\log 3}{\log 2}$ years to treble the population.
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