Question

For $x,y\in N,$ if $3^{2x-y+1}=3^{y-2x+1}-8$ and ${\log }_6|2x^{2}y-x{y}^2|=1+{\log }_{36}\left(xy\right),$ then the absolute value of $(x-y)$ is

Answer 1

$3^{2x-y+1}=3^{y-2x+1}-8$
Let $3^{2x-y}=t$
Then, $3t=\frac{3}{t}-8$
$\Rightarrow 3t^2+8t-3=0$
$\Rightarrow t=\frac{1}{3}$ or $t=-3$ (rejected)
Therefore, $3^{2x-y}=t=3^{-1}$
$\Rightarrow 2x-y=-1\cdots \left(1\right)$

Also, ${\log }_6|2x^{2}y-x{y}^2|=1+{\log }_{36}\left(xy\right)$
$\Rightarrow {\log }_6\left|xy\right(2x-y\left)\right|=1+{\log }_6\sqrt{xy}$
$\Rightarrow {\log }_6\left|xy\right|={\log }_6\left(6\sqrt{xy}\right)$
$\Rightarrow xy=36\cdots \left(2\right)$

On solving $\left(1\right)$ and $\left(2\right),$ we get
$x=4$ and $y=9$
$\therefore |x-y|=5$