Question

The smallest integer satisfying the equation
$\left|{\log }_{\frac{1}{6}}x-1\right|+2=\left|{\log }_{\frac{1}{6}}x-3\right|$ is

Answer 1

Clearly, $x>0$. Put ${\log }_{\frac{1}{6}}x=t$, so that
$|1-t|+2=|3-t|\Rightarrow t\le 1$
$\Rightarrow x\ge \frac{1}{6}$ or $x\le {\left(\frac{1}{6}\right)}^3$
Ans: 1