$\mathrm{if} \frac{\log \mathrm{p}}{2} = \frac{\log \mathrm{q}}{4} = \frac{\log \mathrm{r}}{8} = \mathrm{k} \mathrm{and}, \mathrm{pqr} = 100, \mathrm{k} =$

$$\begin{gathered} \mathrm{if} \frac{\log \mathrm{p}}{2} = \frac{\log \mathrm{q}}{4} = \frac{\log \mathrm{r}}{8} = \mathrm{k} \\ \frac{{\log }_{10}\left(\mathrm{p}\right)}{2} = \mathrm{k} \Rightarrow {\log }_{10}\left(\mathrm{p}\right) = 2\mathrm{k} \Rightarrow \mathrm{p} = {10}^{2\mathrm{k}} \\ \frac{{\log }_{10}\left(\mathrm{q}\right)}{4} = \mathrm{k} \Rightarrow {\log }_{10}\left(\mathrm{q}\right) = 4\mathrm{k} \Rightarrow \mathrm{q} = {10}^{4\mathrm{k}} \\ \frac{{\log }_{10}\left(\mathrm{r}\right)}{8} = \mathrm{k} \Rightarrow {\log }_{10}\left(\mathrm{r}\right) = 8\mathrm{k} \Rightarrow \mathrm{r} = {10}^{8\mathrm{k}} \\ \mathrm{Given} \mathrm{that} \mathrm{pqr} = 100 \\ \Rightarrow {10}^{2\mathrm{k}} .{10}^{4\mathrm{k}}. {10}^{8\mathrm{k}} = {10}^2 \\ \Rightarrow {10}^{\left(2+4+8\right)\mathrm{k}} = {10}^2 \\ \Rightarrow {10}^{14\mathrm{k}} = {10}^2 \\ \Rightarrow 14\mathrm{k} =2 \\ \Rightarrow \mathrm{k} = \frac{2}{14} \\ \Rightarrow \mathrm{k} = \frac{1}{7} \end{gathered}$$