\begin{aligned}
&\lambda=\frac{0.693}{t_{1} / 2}=\frac{0.693}{138 \text { days }}=5.022 \times 10^{-3} dy ^{-1} \\
&A_{t}=A_{o} e^{-\lambda t} \ldots \ldots \ldots .( i )
\end{aligned}
The activity decreasing to \(5 \%\) of hte initial value means \(\frac{A_{t}}{A_{o}} \times 100 \%=15 \%\)
Then, \(\frac{A_{t}}{A_{o}}=0.15\)rom (i)
\(\frac{A_{t}}{A_{0}}=e^{-\lambda t}\)
Therefore,
\(0.15=e^{-5.022 \times 10^{-3} t}\).
Introduce In to both sides
\begin{aligned}
\ln 0.15 &=\ln \left(e^{-5.022 \times 0^{-3} t}\right) \\
\ln 0.15 &=-5.022 \times 10^{-3} t \\
t &=\frac{\ln 0.15}{-5.022 \times 10^{-3}} \\
&=378 \text { days }
\end{aligned}