Question

$\sqrt{{\log }_{2}x}-0.5={\log }_2\sqrt{x}$,
then $x$ is equal to

• A. None of these
• B. $4$
  
• C. $16$
  
• D. $8$
  

Answer 1

The correct option is A

None of these

$\text{Given}:\sqrt{{\log }_{2}x}-0.5={\log }_2\sqrt{x}\Rightarrow \sqrt{{\log }_{2}x}=\frac{1}{2}+{\log }_2{x}^{\frac{1}{2}}\Rightarrow \sqrt{{\log }_{2}x}=\frac{1}{2}+\frac{1}{2}{\log }_{2}x$
On squaring both sides, we get
${\log }_{2}x=\frac{1+2{\log }_{2}x+({\log }_{2}x{)}^2}{4}$
$\Rightarrow 4{\log }_{2}x=1+2{\log }_{2}x+({\log }_{2}x{)}^2$
$\Rightarrow ({\log }_{2}x{)}^2-2{\log }_{2}x+1=0$
$\Rightarrow ({\log }_{2}x-1{)}^2=0$
$\Rightarrow {\log }_{2}x=1$
$\Rightarrow x=2$
Hence, The correct answer is option (d)