Question

The integral value of $x$ satisfying the equation $|{\log }_{\sqrt{3}}x-2|-|{\log }_{3}x-2|=2,$ is

Answer 1

Clearly, $x>0$
$|{\log }_{\sqrt{3}}x-2|-|{\log }_{3}x-2|=2$
$\Rightarrow |2{\log }_{3}x-2|-|{\log }_{3}x-2|=2$
Let ${\log }_{3}x=t$
Then, $2|t-1|-|t-2|=2$

Case $1:$ $t\ge 2$
$2(t-1)-(t-2)=2$
$\Rightarrow t=2$
$\Rightarrow x=9$

Case $2:$ $1<t<2$
$2(t-1)+(t-2)=2$
$\Rightarrow t=2$ Not possible

Case $3:$ $t\le 1$
$-2(t-1)+(t-2)=2$
$\Rightarrow t=-2$
$\Rightarrow x=\frac{1}{9}$