Question

Find the solutions set of ${\log }_{x^2}(\frac{x}{|x|}-x)\ge 0$.

Answer 1

If $x\ge 0$

$\Rightarrow {\log }_{x^2}(\frac{x}{|x|}-x)$

$={\log }_{x^2}(\frac{x}{x}-x)={\log }_{x^2}(1-x)$

$=\frac{1}{2}{\log }_x(1-x)\ge 0$

$\Rightarrow {\log }_x(1-x)\ge 0$

Case-I: $x<1$ and $(1-x)\le 1$

$\Rightarrow x<1$ and $x\ge 0\to$ Possible

Case-II: $x>1$ and $(1-x)\ge 1$

$\Rightarrow x>1$ and $x\le 0\to$ Not Possible

$\therefore xϵ[0,1)$

If $x<0$

$\Rightarrow {\log }_{x^2}(\frac{x}{|x|}-x)={\log }_{x^2}(-1-x)\ge 0$

$\Rightarrow {\log }_{x^2}(-1-x)\ge 0$

Case-I: $x^2>1$ and $(-1-x)\ge 1$

$\Rightarrow x<-1$ and $x\le -2$

(Since $x<0$)

$\therefore xϵ(-\infty ,-2]$

Case-II: $x^2<1$ and $(-1-x)\le 1$

$\Rightarrow -1<x<0$ and $x\ge -2$

(Since $x<0$)

$\therefore xϵ(-1,0)$

$\therefore$ Combining all the above cases, we get

$xϵ(-\infty ,-2]\cup (-1,1)$