Answer
$ t \approx 12.5$ days
Work Step by Step
The exponential decay process can be written as: $ N(t) N_0 e^{-\lambda t}$ ...(1)
But $\lambda =\dfrac{\ln 2}{t_{1/2}}= 0.096$
We need to simply equation (1) to obtain an expression for $ t $ when $ N(t)=0.3 N_0$
Plug in the given data:
$0.3 N_0=N_0 e^{-0.096} t $
$\ln 0.3 =\ln (e^{-0.096t})$
or, $\ln 0.3 =-0.096t $
Thus, $ t \approx 12.5$ days
