# E-8 10'09, (log, (log, x)) = 1 and log, (log, (log, x)) = 0 then 'p' equals (A) rar (B) rg (C) 1 (D)rla

${10}^{\mathrm{log}p\left({\mathrm{log}}_{q}\right(\mathrm{log}x\left)\right)}=1\phantom{\rule{0ex}{0ex}}\Rightarrow {\mathrm{log}}_{p}\left({\mathrm{log}}_{q}\right({\mathrm{log}}_{r}x\left)\right)=0\phantom{\rule{0ex}{0ex}}\Rightarrow {\mathrm{log}}_{q}\left({\mathrm{log}}_{r}x\right)=1\phantom{\rule{0ex}{0ex}}\Rightarrow {\mathrm{log}}_{r}x=q\Rightarrow x={r}^{q}....1\phantom{\rule{0ex}{0ex}}{\mathrm{log}}_{q}\left({\mathrm{log}}_{r}\right({\mathrm{log}}_{p}x\left)\right)=0\phantom{\rule{0ex}{0ex}}\Rightarrow {\mathrm{log}}_{r}\left({\mathrm{log}}_{p}x\right)=1\phantom{\rule{0ex}{0ex}}\Rightarrow {\mathrm{log}}_{p}x=r\phantom{\rule{0ex}{0ex}}\Rightarrow x={p}^{r}.....2\phantom{\rule{0ex}{0ex}}form\left(1\right)2\phantom{\rule{0ex}{0ex}}{r}^{q}={p}^{r}\phantom{\rule{0ex}{0ex}}q\mathrm{log}r=r\mathrm{log}p\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{log}p=\frac{q}{r}\mathrm{log}r=\mathrm{log}{r}^{q/r}\phantom{\rule{0ex}{0ex}}\mathbf{\Rightarrow}\mathit{p}\mathbf{=}{\mathit{r}}^{\mathbf{q}\mathbf{/}\mathbf{r}}\phantom{\rule{0ex}{0ex}}\mathit{O}\mathit{p}\mathit{t}\mathit{i}\mathit{o}\mathit{n}\mathbf{}\mathbf{\left(}\mathbf{1}\mathbf{\right)}$