A radioactive nuclide X of half-life 2 h decays to a nuclide Y of half-life 6 h, which decays to a stable nuclide. At t=0 only X is present. The…

With decay constants $\lambda_X=\frac{\ln2}{2}$ and $\lambda_Y=\frac{\ln2}{6}$ per hour, $N_Y(t)=\frac{\lambda_XN_0}{\lambda_Y-\lambda_X}\left(e^{-\lambda_Xt}-e^{-\lambda_Yt}\right)$. The number of $Y$ is maximum when $\frac{dN_Y}{dt}=0$, i.e. when the rate of production equals the rate of decay: $\lambda_XN_X=\lambda_YN_Y$. Differentiating the expression gives $\lambda_Xe^{-\lambda_Xt}=\lambda_Ye^{-\lambda_Yt}$, so $t_{max}=\frac{\ln(\lambda_X/\lambda_Y)}{\lambda_X-\lambda_Y}=\frac{\ln3}{\ln2\,(\frac12-\frac16)}=\frac{3\ln3}{\ln2}=3\log_23\approx4.75$ h. At this instant the activity of $Y$ equals that of $X$, which is a useful check: $\lambda_XN_X=\lambda_YN_Y$ holds exactly at $t_{max}$. Because $\lambda_X>\lambda_Y$ the logarithm is positive and the maximum occurs at a finite time, about $4.75$ hours, well after the first half-life of $X$ and before the half-life of $Y$ has elapsed once, as expected.