Question

Solve the following inequality:
$\frac{{\log }_2(x+1)}{x-1}>0$

Answer 1

either

$lo{g}_2(x+1)>0andx-1>0$

$x+1>1andx>1$

$x>0andx>1$

$therefore,x>1$

$lo{g}_2(x+1)<0andx-1<0$

$0<x+1<1andx<1$

$therefore,xϵ(-1,0)$

Hence,we can say that

$xϵ(-1,0)\cup (1,\infty )$