Question

The rate of growth of a population is proportional to the number present. If the population of a city doubled in the past 25 years, and the present population is 100000, when will the city have a population of 500000?

Answer 1

Let the original population be N and the population at any time t be P.
Given:-

\[\Rightarrow \frac{dP}{dt} = aP\]
\[ \Rightarrow \frac{dP}{P} = adt\]
\[ \Rightarrow \log\left| P \right| = \text{ at }+ C . . . . . . . . . . \left( 1 \right)\]
Now, 
\[P = N\text{ at }t = 0\]
\[\text{ Putting }P = N\text{ and }t = 0\text{ in }\left( 1 \right), \text{ we get }\]
\[\log\left| N \right| = C\]
\[\text{ Putting }C = \log\left| N \right| \text{ in }\left( 1 \right), \text{ we get }\]
\[\log\left| P \right| =\text{ at }+ \log\left| N \right|\]
\[ \Rightarrow \log\left| \frac{P}{N} \right| = at . . . . . . . . . \left( 2 \right)\]
According to the question,
\[\log\left| \frac{2N}{N} \right| = 25a\]
\[ \Rightarrow a = \frac{1}{25}\log\left| 2 \right| = \frac{1}{25} \times 0 . 6931 = 0 . 0277\]
\[\text{ Putting }a = 0 . 0277\text{ in }\left( 2 \right),\text{ we get }\]
\[\log\left| \frac{P}{N} \right| = 0 . 0277t . . . . . . . . \left( 3 \right)\]
\[\text{ For }P = 500000\text{ and }N = 100000: \]
\[\log\left| \frac{500000}{100000} \right| = 0 . 0277t\]
\[ \Rightarrow t = \frac{\log 5}{0 . 0277} = \frac{1 . 609}{0 . 0277} = 58 . 08\text{ years }=\text{ Approximately 58 years}\]