A charged capacitor of capacitance C is discharging through a...

The charge on the capacitor at time $t$ in a discharging circuit is given by $q = {q}_{0}e^{- t / τ}$

where, $τ = RC =$ time constant
$∴$ & & & & & & $q = {q}_{0}e^{- t / RC}$

when, $q = \dfrac{q0}{2}$

then $\dfrac{q0}{2} = {q}_{0}e^{- t / RC}$

$\dfrac{1}{2} = e^{- t / RC}$

Taking natural log on both side $\log_{\mathrm{e}}\left(\dfrac{1}{2}\right) = \log_{0}e^{- t / RC}$

$- \log_{\mathrm{e}}2 = - \dfrac{t}{RC}\log_{\mathrm{e}}e$

$⇒ t = RC\log_{\mathrm{e}}2$