Rate of increase of population is proportional to Number of people living presently

After 2 years, Population has doubled

After 3 years, Population \(\displaystyle={20000}\)

Let P be population in the country.

Thus, according to the question:

\(\displaystyle{\frac{{{d}{P}}}{{{\left.{d}{t}\right.}}}}\infty{P}\)

\(\displaystyle{\frac{{{d}{P}}}{{{\left.{d}{t}\right.}}}}={k}{P}\), where k is any constant.

\(\displaystyle{\frac{{{d}{P}}}{{{P}}}}={k}.{\left.{d}{t}\right.}\)

\(\displaystyle\int{\frac{{{d}{P}}}{{{P}}}}=\int{k}.{\left.{d}{t}\right.}\)

\(\displaystyle{\ln{{P}}}={k}.{t}+{C}\)

When \(\displaystyle{t}={0}\) years, \(\displaystyle{P}={P}_{{{0}}}\), where \(\displaystyle{P}_{{{0}}}\) is the initial population.

Thus, \(\displaystyle{{\ln{{P}}}_{{{0}}}=}{0}+{C}\)

\(\displaystyle{C}={{\ln{{P}}}_{{{0}}}}\)

\(\displaystyle{\ln{{P}}}={k}{t}+{{\ln{{P}}}_{{{0}}}}\)

\(\displaystyle{\ln{{P}}}-{{\ln{{P}}}_{{{0}}}=}{k}{t}\)

\(\displaystyle{\ln{{\left({\frac{{{P}}}{{{P}_{{{0}}}}}}\right)}}}={k}{t}\)

When \(\displaystyle{t}={2}\) years, \(\displaystyle{P}={2}{P}_{{{0}}}\)

\(\displaystyle{\ln{{\left({\frac{{{2}{P}_{{{0}}}}}{{{P}_{{{0}}}}}}\right)}}}={k}{x}{2}\)

\(\displaystyle{k}={\frac{{{\ln{{2}}}}}{{{2}}}}\)

\(\displaystyle{\ln{{\left({\frac{{{P}}}{{{P}_{{{0}}}}}}\right)}}}={\frac{{{\ln{{2}}}}}{{{2}}}}.{t}\)

When \(\displaystyle{t}={3}\) years, \(\displaystyle{P}={20000}\)

\(\displaystyle{\ln{{\left({\frac{{{20000}}}{{{P}_{{{0}}}}}}\right)}}}={\frac{{{\ln{{2}}}}}{{{2}}}}{x}{3}\)

\(\displaystyle{\ln{{\left({\frac{{{20000}}}{{{P}_{{{0}}}}}}\right)}}}={1.0397}\)

\(\displaystyle{\frac{{{20000}}}{{{P}_{{{0}}}}}}={e}^{{{1.0397}}}\)

\(\displaystyle{P}_{{{0}}}={\frac{{{20000}}}{{{e}^{{{1.0397}}}}}}={7071.2}\)

\(\displaystyle{P}_{{{0}}}={7072}\)

So, initial population, \(\displaystyle{P}_{{{0}}}={7072}\)