Question

Solve the following inequality:
${\log }_{1/3}x>{\log }_{x}3-\frac{5}{2}$

Answer 1

$-\frac{\log x}{\log 3}>\frac{\log 3}{\log x}-\frac{5}{2}\frac{\log x}{\log 3}+\frac{\log 3}{\log x}-\frac{5}{2}<02(\log x{)}^2-5\log 3\log x+2(\log 3{)}^2<02(\log x{)}^2-4\log 3\log x-\log 3\log x+2(\log 3{)}^2<02\log x(\log x-2\log 3)-\log 3(\log x-2\log 3)<0(\log x-2\log 3)(2\log x-\log 3)<0x\in (\sqrt{3},9)$

Also for, $x\in (0,1){\log }_{1/3}x>0$

and ${\log }_{x}3-\frac{5}{2}<0$

Hence,

${\log }_{1/3}x>{\log }_{x}3-\frac{5}{2}$ for $x\in (0,1)$

Taking union we get, $x\in (0,1)\cup (\sqrt{3},9)$