Question

Solve the system of equation:
${\log }_{a}x{\log }_{a}xyz=48$
${\log }_{a}y{\log }_{a}xyz=12$
${\log }_{a}z{\log }_{a}xyz=84$

Answer 1

${\log }_{a}x{\log }_{a}xyz=48$(equation $1$)

${\log }_{a}y{\log }_{a}xyz=12$(equation $2$)

${\log }_{a}z{\log }_{a}xyz=84$(equation $3$)

Adding euation $1,2$ and $3$:

$\Rightarrow {\log }_a\left(xyz\right)({\log }_{a}x+{\log }_{a}y+{\log }_{a}z)=144$

$\Rightarrow {\log }_a\left(xyz\right){\log }_a\left(xyz\right)=144$

$\Rightarrow {\left({\log }_a\left(xyz\right)\right)}^2=144$

$\Rightarrow {\log }_{a}xyz=+12$

$\Rightarrow xyz=a^{12}$

And $\Rightarrow {\log }_{a}xyz=-12$

$\Rightarrow xyz=a^{-12}$

So,

$\Rightarrow xyz=a^{12},a^{-12}$(equation $4$)

equation $1-4\times$ equation $2$

Then,

$\Rightarrow {\log }_{a}x{\log }_{a}xyz-4{\log }_{a}y{\log }_{a}xyz=0$

$\Rightarrow {\log }_{a}xyz({\log }_{a}x-4{\log }_{a}y)=0$

$\Rightarrow {\log }_{a}xyz\ne 0$

And, $\Rightarrow {\log }_{a}x=4{\log }_{a}y$

$\Rightarrow x=y^4$

equation$1-\frac{7}{4}\times$equation$2$

$\Rightarrow {\log }_{a}z{\log }_{a}xyz-\frac{7}{4}{\log }_{a}x{\log }_{a}xyz=0$

$\Rightarrow {\log }_{a}xyz({\log }_{a}z-\frac{7}{4}{\log }_{a}x)=0$

$\Rightarrow {\log }_{a}xyz\ne 0$

And, $\Rightarrow {\log }_{a}z=\frac{7}{4}{\log }_{a}x$

$\Rightarrow z=x^{7/4}$

From equation $4$

$\Rightarrow xyz=a^{12}$

$\Rightarrow \left(x\right)\left(x{)}^{1/4}\right(x{)}^{7/4}=a^{12}$

$\Rightarrow x^3=a^{12}$

$\Rightarrow x=a^4$

So, $y=a,z=a^7$

or,

$\Rightarrow xyz=a^{-12}$

$\Rightarrow \left(x\right)\left(x{)}^{1/4}\right(x{)}^{7/4}=a^{-12}$

$\Rightarrow x^3=a^{-12}$

$\Rightarrow x=a^{-4}$

So, $y=a^{-1},z=a^{-7}$

Therefore, $\left(xyz\right):(a^4,a,a^7)$ or $(a^{-4},a^{-1},a^{-7})$