Question

Solve the following inequality:
${\left(\frac{1}{2}\right)}^{{\log }_2(x^2-1)}>1$

Answer 1

$(\frac{1}{2}{)}^{{log}_2(x^2-1)}>1\Rightarrow (x^2-1{)}^{{log}_2\frac{1}{2}}>1\Rightarrow {(x^2-1)}^{-1}>1\Rightarrow \frac{1}{x^2-1}-1>0\Rightarrow \frac{2-x^2}{x^2-1}>0\Rightarrow \frac{x^2-2}{x^2-1}<0$

$x^2-2>0$ and $x^2-1<0$ or $x^2-2<0$ and $x^2-1>0$

$x\in (-\infty ,-\sqrt{2})\cup (\sqrt{2},\infty )$ and $x\in (-1,1)$ or $x\in (-\sqrt{2},\sqrt{2})$ and $x\in (-\infty ,-1)\cup (1,\infty ).$

$\therefore x\in (-\sqrt{2},-1)\cup (1,\sqrt{2}).$