Question

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

Answer 1

$lo{g}_{x^2}[\frac{x}{|x|}-x]\ge 0$

for any log form such as as ${\log }_a$ b

${\log }_{x}b\ge 0$ implies $b\ge 1$....(1)

${\log }_{x^2}(\frac{x}{|x|}-x)=\left\{\begin{array}{cc}lo{g}_{x^2}(1-x)& x\ge 0\\ {\log }_{x^2}(-1-x)& x<0\end{array}\right\}$...(2)

Here since base of log is $x^2$ , it is always positive.

Using result of equation (1)

$1-x\ge 1\Rightarrow x\le 0$ which contradict $x\ge 0$ is first past of equation (2)

Also, $(-1-x)\ge 1\Rightarrow x\le -2$ and since the domain is $x<0$ The set of volume of $x$ which satisfy the given inequality $=(-\infty ,-2)$