Question

Solve the following inequality:
${\left(\frac{1}{3}\right)}^{\frac{|x+2|}{2-|x|}}>9$

Answer 1

$(\frac{1}{3}{)}^{\frac{|x+2|}{2-\left|x\right|}}>3^2$

$3^{\frac{|x+2|}{\left|x\right|+2}}>3^2$

$\frac{|x+2|}{\left|x\right|-2}>2$

$\frac{|x+2|-2(\left|x\right|-2)}{\left|x\right|-2}>0$

$\frac{|x+2|-2\left|x\right|+4}{\left|x\right|-2}>0$

$\left(i\right)whenx>0$

$\frac{x+2-2x+4}{x-2}>0$

$\frac{x-6}{x-2}<0,x\epsilon (2,6)-\left(1\right)$

$\left(ii\right)whenx<-2$

$\frac{-x-2+2x+4}{-x-2}>0$

$\frac{x+2}{-(x+2)}>0$

$1<0\left(notpossible\right)$

$x\epsilon \left\{\right\}-\left(2\right)$

$\left(iii\right)when-2<x<0$

$\frac{x+2-2(-x-2)}{-x-2}>0$

$\frac{x+2+2x+4}{x+2}<0$

$\frac{x+2}{x+2}<0\left(notpossible\right)$

$x\epsilon \left\{\right\}-\left(3\right)$

$from\left(1\right),\left(2\right)\&\left(3\right),x\epsilon (2,6)$