Given:

\(p(0)=10\) million ... at \( t=0\), 1980

\(p(10)=75\) million ... at \(t=10\) 1990

So, \(p(0)=p_{0}=10\) million

\(\displaystyle\Rightarrow{p}{\left({t}\right)}={p}_{{0}}{e}^{{{k}{t}}}\)

\(\displaystyle\Rightarrow{p}{\left({t}\right)}={10}{e}^{{{k}{t}}}\)

when \(t=10\), (1990) \(\displaystyle\Rightarrow{p}{\left({10}\right)}={75}\)

\(\displaystyle\Rightarrow{p}{\left({10}\right)}={10}{e}^{{{10}{k}}}\Rightarrow{75}={10}{e}^{{{10}{k}}}\)

\(\displaystyle\Rightarrow{e}^{{{10}{k}}}={\frac{{{75}}}{{{10}}}}\)

\(\displaystyle\Rightarrow{e}^{{{10}{k}}}={7.5}\)

Take log on both sides

\(\displaystyle\Rightarrow{10}{k}={{\log}_{{e}}{\left({7.5}\right)}}\)

\(\displaystyle{10}{k}={2.02}\)

\(\displaystyle{k}={\frac{{{2.02}}}{{{10}}}}\)

\(\displaystyle{k}={0.202}\)

\(\displaystyle\Rightarrow{k}={0.2}\)

\(\displaystyle{p}{\left({t}\right)}={10}{e}^{{{0}\cdot{2}{t}}}\)

predicted population in year 2000

\(\displaystyle\Rightarrow{t}={20}\)

\(\displaystyle\Rightarrow{p}{\left({20}\right)}={10}{e}^{{{0}\cdot{2}{\left({20}\right)}}}\)

\(\displaystyle\Rightarrow{p}{\left({20}\right)}={10}{e}^{{4}}\)

\(\displaystyle\Rightarrow{p}{\left({20}\right)}={545.98}\approx{546}\)

Population in 2000 \(p(20)= 546\) million

we, \(\displaystyle{p}{\left({t}\right)}={10}{e}^{{{0}\cdot{2}{t}}}\)

Now,

The population during doubling time \(\displaystyle{p}{\left({t}\right)}={2}\times{10}\) at \(t=0\)

\(\displaystyle{p}{\left({t}\right)}={20}\)

\(\displaystyle\Rightarrow{20}={100}{e}^{{{0}\cdot{2}{t}}}\)

\(\displaystyle\Rightarrow{e}^{{{0}\cdot{2}{t}}}={2}\)

\(\displaystyle{0}\cdot{2}{t}={\ln{{\left({2}\right)}}}\)

\(\displaystyle{0}\cdot{2}{t}={0.69}\)

\(\displaystyle{t}={\frac{{{0.69}}}{{{0.2}}}}\)

\(\displaystyle{t}={3.47}\) doubling time