Question

The solution set of the inequality ${\log }_{2}x\le \frac{2}{{\log }_{2}x-1}$ is

• A. $(\frac{1}{2},4)-\left\{2\right\}$
• B. $(0,4]-\left\{2\right\}$
• C. $(0,\frac{1}{2})$
• D. $(0,\frac{1}{2}]\cup (2,4]$

Answer 1

The correct option is D $(0,\frac{1}{2}]\cup (2,4]$

${\log }_{2}x\le \frac{2}{{\log }_{2}x-1}$
Clearly, $x>0$
and ${\log }_{2}x\ne 1\Rightarrow x\ne 2$

Now, ${\log }_{2}x-\frac{2}{{\log }_{2}x-1}\le 0$
$\Rightarrow \frac{({\log }_{2}x{)}^2-{\log }_{2}x-2}{{\log }_{2}x-1}\le 0$
$\Rightarrow \frac{({\log }_{2}x-2)({\log }_{2}x+1)}{{\log }_{2}x-1}\le 0$
By using wavy curve method, we get
$\Rightarrow {\log }_{2}x\in (-\infty ,-1]\cup (1,2]$
$\Rightarrow x\in (0,\frac{1}{2}]\cup (2,4]$