The activity of a radioactive sample is measured as N${}_0$ counts per minute at t = 0 and N${}_0$/e counts per minute at t = 5 minutes. The time (in minutes) at which the activity reduces to half its value is

According to question, at $t=5$ $min$

$N=\frac{N_o}{e}$

Therefore, equation of nuclei left undecayed after time, $t$ is given by,

$N=N_o{\left(\frac{1}{2}\right)}^{t/T_{1/2}}$

when $t=5$ $min$, $N=\frac{N_o}{e}$

$⟹\frac{N_o}{e}=N_o{\left(\frac{1}{2}\right)}^{5/T_{1/2}}$

$⟹e^{-1}={\left(\frac{1}{2}\right)}^{5/T_{1/2}}$

Calling log both sides,

$⟹-\ln (e)=\frac{5}{T_{1/2}}\ln \left(\frac{1}{2}\right)$  

And as we know $ln(e)=1$

$⟹T_{1/2}=-5\ln (1/2)$

$⟹T_{1/2}=-5(-\ln 2)$

$⟹T_{1/2}=5\ln 2$

$⟹T_{1/2}=5{\log }_{e}2$