Question

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

Answer 1

$\begin{array}{c}{\log }_{2}x\le \frac{2}{{\log }_{2}x-1}\\ \Rightarrow {\left({\log }_{2}x\right)}^2-{\log }_{2}x\le 2\\ Let{\log }_{2}x=t\\ \Rightarrow t^2-t-2\le 0\\ \Rightarrow t^2-2t+t-2\le 0\\ \Rightarrow t\left(t-2\right)+1\left(t-2\right)\le 0\\ \Rightarrow \left(t+1\right)\left(t-2\right)\le 0\\ {\log }_{2}x=-1\\ \Rightarrow 2^{-1}\le x\\ x\ge \frac{1}{2}\\ and\\ \Rightarrow {\log }_{2}x\ge 2\\ \Rightarrow 2^2\ge x\\ x\le 4\\ x\in {\left[2,4\right]}^2\end{array}$