Question

Solve the following inequality:
$\frac{1}{1+\log x}+\frac{1}{1-\log x}>2$

Answer 1

$\frac{1}{1+\log x}+\frac{1}{1-\log x}>2⟹\frac{2}{1-(\log x{)}^2}>2$

$\frac{2}{1-(\log x{)}^2}-2>0⟹\frac{2-2+2(\log x{)}^2}{1-(\log x{)}^2}>0$

$⟹1-(\log x{)}^2>0⟹\log x\in (-1,1)⟹x\in (\frac{1}{10},10)$