Question

Solve the following inequality:
$\frac{x^2-4}{{\log }_{\frac{1}{3}}(x^2-1)}<0$

Answer 1

$either{x}^2-4>0andlo{g}_3(x^2-1)>0xϵ(-\infty ,-2)\cup (2,\infty )and{x}^2-1>1x^2-2>0xϵ(-\infty ,-\sqrt{2})\cup (\sqrt{2},\infty )So,xϵ(-\infty ,-2)\cup (2,\infty )therefore,or{x}^2-4<0{\log }_3(x^2-1)<0xϵ(-2,2)and{\log }_3(x^2-1)<0and0<x^2-1<1x^2-1>0xϵ(-\infty ,-1)\cup (1,\infty )and{x}^2-1<0x^2-2<0xϵ(-\sqrt{2},\sqrt{2})xϵ(-\infty ,-2)\cup (-\sqrt{2},-1)\cup (1,\sqrt{2})\cup (2,\infty )$