Question

Solve this :


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

Answer 1

Hi,

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