If the population of a country doubles in 50 years, in how many years will it triple  under the assumption that the rate of increase is proportional to the number of  people.
 

Let x denote the population at a time t in years .
then $\frac{\mathrm{dx}}{\mathrm{dt}}\propto \mathrm{x}\Rightarrow \frac{\mathrm{dx}}{\mathrm{dt}}=\mathrm{kx},$
Where K is constant of proportionality
Solving $\frac{dx}{dt}=kx$, we get $\int \frac{\mathrm{dx}}{\mathrm{x}}=\int \mathrm{kdt}$
$\Rightarrow \log \mathrm{x}=\mathrm{kt}+\mathrm{c}\Rightarrow \mathrm{x}={\mathrm{e}}^{\mathrm{kt}+\mathrm{c}}\Rightarrow \mathrm{x}={\mathrm{x}}_0{\mathrm{e}}^{\mathrm{kt}}$
where x₀ is the population at time t = 0.
Since it doubles in 50 years, at t = 50, we must have x = 2x₀.
Hence$2{\mathrm{x}}_0={\mathrm{x}}_0{\mathrm{e}}^{\mathrm{k}0\mathrm{k}}\Rightarrow 50\mathrm{k}=\log 2$
$\Rightarrow \mathrm{k}=\frac{\log 2}{50}$, so that $\mathrm{x}={\mathrm{x}}_0{\mathrm{e}}^{\frac{\log 2}{50}\mathrm{t}}$
to find t, when it triples i.e, $\mathrm{x}=3{\mathrm{x}}_0$
$\begin{array}{l}\Rightarrow 3{\mathrm{x}}_0={\mathrm{x}}_0{\mathrm{e}}^{\frac{\log 2}{50}\mathrm{t}}\Rightarrow 3={\mathrm{e}}^{\frac{\log 2}{50}\mathrm{t}}\\ \Rightarrow \frac{\log 2}{50}\mathrm{t}=\log 3\Rightarrow \mathrm{t}=50\frac{\log 3}{\log 2}=80\text{ years }\end{array}$