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Answer: $58$ years

Given:  Present Population $= 1,00,000$

To find: When the city will have a population of $5,00,000$

Hint: The population of city increase in time i.e. $\frac{d P}{d t} \propto P$ and then find the equation using integration.

Solution: The present population is $1,00,000$ and the population of a city doubled in the past $25$ years

Let $P$ be the population at any time $t$

Then, $\frac{d P}{d t} \propto P$

\begin{aligned} &\frac{d P}{d t}=k P \\\\ &\frac{d P}{d t}=k d t \end{aligned}

Integrating on both sides we get,

\begin{aligned} &=>20 \int \frac{d P}{P}=\int k d t \\\\ &=>\log P=k t+C \ldots(i) \end{aligned}

At $t=0$ we take $P=P_{0}$

Then,

$=>\log P_{0}=k \times 0+C$                [Putting $t=0$  and $P=P_{0}$  in equation (i)]

$=>C=\log P_{0}$

Putting $C=\log P_{0}$  in equation ($i$) we get

\begin{aligned} &=>\log P=k t+\log P_{0} \\\\ &=>\log P-\log P_{0}=k t \\\\ &=>\log \frac{P}{p_{0}}=k t \ldots(i i) \end{aligned}

When $P=2P_{0}$  at $t=25$ we have

\begin{aligned} &\log \left(\frac{2 P_{o}}{P}\right)=k t \\\\ &=>\log 2=25 k \end{aligned}

Putting $k=\frac{\log 2}{25}$  in equation (ii) we get

$=\log \left(\frac{p}{p_{0}}\right)=\left(\frac{\log 2}{25}\right) t$

We assume that $t_{1}$ be the time take for the population to become $5,00,000$ from $1,00,000$

Then,$\log g\left(\frac{5,00,000}{1,00,000}\right)=\left(\frac{\log 2}{25}\right) t_{1}$

\begin{aligned} &=\log 5=\left(\frac{\log 2}{25}\right) t_{1} \\\\ &=25 \log 5=\log 2 t_{1} \\\\ &=t_{1}=25\left(\frac{\log \log 5}{\log \log 2}\right) \\\\ &=t_{1}=25\left(\frac{1.609}{0.6931}\right) \end{aligned}        $[\log \log 5=1.609, \log \log 2=0.6931]$

$=t_{1}=58.08 \text { Years }$

Therefore the required time is $58$ years (approximate).

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