Solve this :


$$\begin{gathered} \mathit{Q}\mathbf{.}\mathbf{10}\mathbf{.}\mathbf{ }\mathbf{ } Solve the system of equations : \\ {\log }_a x {\log }_a \left(xyz\right) = 48 \\ {\log }_a y {\log }_a \left(xyz\right) = 12, a > 0, a \ne 1. \\ {\log }_a z {\log }_a \left(xyz\right) = 84 \end{gathered}$$

$$\begin{gathered} \left({\log }_{\mathrm{a}}\mathrm{x}\right)\times {\log }_{\mathrm{a}}\left(\mathrm{xyz}\right)=48....\left(1\right) \\ \left({\log }_{\mathrm{a}}\mathrm{y}\right)\times {\log }_{\mathrm{a}}\left(\mathrm{xyz}\right)=12....\left(2\right) \\ \left({\log }_{\mathrm{a}}\mathrm{z}\right)\times {\log }_{\mathrm{a}}\left(\mathrm{xyz}\right)=84....\left(3\right) \\ \mathrm{Add} \mathrm{the} \mathrm{three} \mathrm{equation} \mathrm{we} \mathrm{get} \\ {\log }_{\mathrm{a}}\left(\mathrm{xyz}\right)\left[{\log }_{\mathrm{a}}\mathrm{x}+{\log }_{\mathrm{a}}\mathrm{y}+{\log }_{\mathrm{a}}\mathrm{z}\right]= 144 \\ {\log }_{\mathrm{a}}\left(\mathrm{xyz}\right)\times {\log }_{\mathrm{a}}\left(\mathrm{xyz}\right)= 144 \\ {\log }_{\mathrm{a}}\left(\mathrm{xyz}\right)= 12 \mathrm{or} -12 \\ \mathrm{now} \mathrm{case} 1, \mathrm{when} {\log }_{\mathrm{a}}\left(\mathrm{xyz}\right)= 12 \\ \mathrm{then} \mathrm{from} \mathrm{equation} 1 , 2 ,3 \\ {\log }_{\mathrm{a}}\mathrm{x}=\frac{48}{12}= 4 \mathrm{so} \mathrm{x} = {\mathrm{a}}^4 \\ {\log }_{\mathrm{a}}\mathrm{y}=\frac{12}{12}=1 \mathrm{so} \mathrm{y} = {\mathrm{a}}^1 \\ {\log }_{\mathrm{a}}\mathrm{z}=\frac{84}{12}= 7\mathrm{so} \mathrm{z} = {\mathrm{a}}^7 \\ \mathrm{now} \mathrm{case} 2, \mathrm{when} {\log }_{\mathrm{a}}\left(\mathrm{xyz}\right)=- 12 \\ \mathrm{then} \mathrm{from} \mathrm{equation} 1 , 2 ,3 \\ {\log }_{\mathrm{a}}\mathrm{x}=\frac{48}{-12}=- 4 \mathrm{so} \mathrm{x} = {\mathrm{a}}^{-4} \\ {\log }_{\mathrm{a}}\mathrm{y}=\frac{12}{-12}=-1 \mathrm{so} \mathrm{y} = {\mathrm{a}}^{-1} \\ {\log }_{\mathrm{a}}\mathrm{z}=\frac{84}{-12}= -7\mathrm{so} \mathrm{z} = {\mathrm{a}}^{-7} \end{gathered}$$