The rate of decay of mass of certain substance at time tis proportional to the tt that instant. The time during which the original mass of \(m_{0}\) gram will be leftht \(m_{\text {, gram is }} \quad(k\) is constant of proportionality

Let \(m\) be the initial mass of substance at time \(t\).
\(
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\therefore \frac{\mathrm{dm}}{\mathrm{dt}} \propto \mathrm{km} \Rightarrow \frac{\mathrm{dm}}{\mathrm{dt}}=-\mathrm{km} \quad \Rightarrow \int \frac{\mathrm{dm}}{\mathrm{m}}=\int-\mathrm{k} \mathrm{dt} \\
\therefore \log \mathrm{m}=-\mathrm{kt}+\mathrm{c}....(1) \\
\text { When } \mathrm{t}=0, \mathrm{~m}=\mathrm{m}_{0} \text { we get } \\
\log \mathrm{m}_{0}=0+\mathrm{c} \Rightarrow \mathrm{c}=\log \mathrm{m}_{0} \\
\text { From }(1), \log \left(\frac{\mathrm{m}}{\mathrm{m}_{0}}\right)=-\mathrm{kt} \Rightarrow \mathrm{t}=\frac{-1}{\mathrm{k}} \log \left(\frac{\mathrm{m}}{\mathrm{m}_{0}}\right)
\end{array}
\)
\(\text {When } m=m_{1} \text {, we get} \)
\( t =\frac{-1}{k} \log \left(\frac{m_{1}}{m_{0}}\right) \Rightarrow t=\frac{1}{k} \log \left(\frac{m_{0}}{m_{1}}\right)\)