The rate of decay of mass of a certain substance at time 't' is proportional to the mass at that instant. The time during which the original mass of \(m_{0}\) gm. will be left to \(m_{1} \mathrm{gm}\). is (K is constant of proportionality)

Given \(\frac{\mathrm{dm}}{\mathrm{dt}} \propto \mathrm{m} \Rightarrow \frac{\mathrm{dm}}{\mathrm{dt}}=-\mathrm{Km} \Rightarrow \int \frac{\mathrm{dm}}{\mathrm{m}}=\int-\mathrm{Kdt}\)
\(\therefore \log \mathrm{m}=-\mathrm{Kt}+\mathrm{c}\)
When \(\mathrm{t}=0, \mathrm{~m}=\mathrm{m}_{0}\)
\(\therefore \log \mathrm{m}_{0}=0+\mathrm{c} \Rightarrow \mathrm{c}=\log \mathrm{m}_{0}\)
\(
\log \mathrm{m}=-\mathrm{Kt}+\log \mathrm{m}_{0}
\)
\(\therefore \log \left(\frac{\mathrm{m}}{\mathrm{m}_{0}}\right)=-\mathrm{Kt}\)
When \(\mathrm{m}=\mathrm{m}_{1}\), we get
\(
\log \left(\frac{\mathrm{m}_{1}}{\mathrm{~m}_{0}}\right)=-\mathrm{Kt}
\)
\(\mathrm{t}=\frac{-1}{\mathrm{~K}} \log \left(\frac{\mathrm{m}_{1}}{\mathrm{~m}_{0}}\right)=\frac{1}{\mathrm{~K}} \log \left(\frac{\mathrm{m}_{0}}{\mathrm{~m}_{1}}\right)\)