Question

Solve the following inequality:
${\log }_{\frac{x+6}{3}}\left({\log }_2\frac{x-1}{2+x}\right)>0$

Answer 1

$\frac{x-1}{2+x}>0(x-1)(x+2)>0x\ne -2x\in (-\infty ,-2)\cup (1,\infty )$

and $\frac{x+6}{3}>0x>-6x\in (-6,\infty )$

Taking intersection, we get $x\in (-6,-2)\cup (1,\infty )$

For $x\in (-6,-3),\log \frac{x+6}{3}<0$

Hence,

${\log }_2\frac{x-1}{2+x}<0\frac{x-1}{2+x}<1(x-1)(2+x)<(x+2{)}^2{x}^2+x-2<x^2+4x+43x+6>0x>-2\Rightarrow x\in (-2,\infty )$

Hence, $x\in \varphi$

For $x\in (-3,-2)\cup (1,\infty )$, $\log \frac{x+6}{3}>0$

${\log }_2\frac{x-1}{2+x}>0\frac{x-1}{2+x}>1(x-1)(2+x)>(x+2{)}^2{x}^2+x-2>x^2+4x+43x+6<0x<-2\Rightarrow x\in (-\infty ,-2)$

Hence, $x\in (-3,-2)$

Taking union, we get$x\in (-3,-2)$