Question

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

Answer 1

${\log }_{\sqrt{3}}(x-2)-{\log }_3(x-2)=2$

$\frac{{\log }_3(x-2)}{{\log }_3\sqrt{3}}-{\log }_3(x-2)=2$

$|\because {\log }_3\sqrt{3}=\frac{1}{2}{\log }_{3}3=\frac{1}{2}$

$2{\log }_3(x-2)-{\log }_3(x-2)={\log }_3\left(3^2\right)$ $|\because 2=2{\log }_{3}3=2\times 1$

${\log }_3(x-2)={\log }_3\left(9\right)$

$x-2=9$

$\therefore x=11$