Question

Solve the following inequality:
$\frac{1-{\log }_{4}x}{1+{\log }_{2}x}⩽\frac{1}{2}$

Answer 1

$\frac{1-{\log }_{4}x}{(1+{\log }_{2}x)}\le \frac{1}{2}$

$\frac{1-\frac{1}{2}{\log }_{2}x}{1+{\log }_{2}x}\le \frac{1}{2}$

$\frac{1}{2}\frac{(2-{\log }_{2}x)}{(1+{\log }_{2}x)}\le \frac{1}{2}$

$\frac{(2-{\log }_{2}x)}{(1+{\log }_{2}x)}\le 1$

$(2-{\log }_{2}x)(1+{\log }_{2}x)\le {(1+{\log }_{2}x)}^2$

$2+{\log }_{2}x-{\left({\log }_{2}x\right)}^2\le {\left({\log }_{2}x\right)}^2+2{\log }_{2}x+1$

$2{\left({\log }_{2}x\right)}^2\left({\log }_{2}x\right)-1\ge 0$

$2{\left({\log }_{2}x\right)}^2+2{\log }_{2}x-{\log }_{2}x\ge 0$

$({\log }_{2}x+1)(2{\log }_{2}x-1)\ge 0$

${\log }_{2}xϵ[-1,\frac{1}{2}]$

$xϵ[\frac{1}{2},\sqrt{2}]$