Answer
Proof.
Work Step by Step
Now, suppose at any time $T$, $P\left( T \right)$ is half of ${{P}_{0}}$, so put $P\left( T \right)=\frac{1}{2}{{P}_{0}}$ in the exponential decay function:
$\begin{align}
& P\left( T \right)={{P}_{0}}{{e}^{-kT}} \\
& \frac{1}{2}{{P}_{0}}={{P}_{0}}{{e}^{-kT}} \\
& \frac{1}{2}={{e}^{-kT}}
\end{align}$
$\begin{align}
& \frac{1}{2}={{e}^{-kT}} \\
& \ln \frac{1}{2}=\ln {{e}^{-kT}} \\
& \ln \frac{1}{2}=-kT \\
& \frac{\ln \frac{1}{2}}{-k}=T
\end{align}$
Further simplified,
$\begin{align}
& \frac{\ln \frac{1}{2}}{-k}=T \\
& \frac{\ln 1-\ln 2}{-k}=T \\
& \frac{0-\ln 2}{-k}=T \\
& \frac{\ln 2}{k}=T
\end{align}$
