Question

Solve the following equation:
$\frac{1}{6}{\log }_2(x-2)-\frac{1}{3}={\log }_{1/8}\sqrt{3x-5}$

Answer 1

$\frac{1}{6}{log}_2(x-2)-\frac{1}{3}=lo{g}_{\frac{1}{8}}\sqrt{3x-5}\Rightarrow {log}_2(x-2)-2=-2{log}_2\sqrt{3x-5}\Rightarrow {log}_2\left(\frac{x-2}{4}\right)={log}_2\frac{1}{3x-5}\Rightarrow \frac{x-2}{4}=\frac{1}{3x-5}\Rightarrow {3x}^2-6x-5x+10=4\Rightarrow {3x}^2-11x+6=0\Rightarrow x=3,\frac{2}{3}$

If $x=\frac{2}{3},x-2<0$ and $3x-5<0$

Therefore, $x\ne \frac{2}{3}$

Therefore, $x=3$ is only solution.