Question

If $\sqrt{{\log }_{2}x}-0.5={\log }_2\sqrt{x}$, then the value of $x$ is

• A. an odd integer.
• B. a prime number.
• C. a composite number.
• D. an irrational number.

Answer 1

The correct option is B a prime number.

$\sqrt{{\log }_{2}x}=0.5+{\log }_2\sqrt{x}$, where $x>0$
For above equation to be defined, ${\log }_{2}x>0\Rightarrow x>1$
$\Rightarrow \sqrt{{\log }_{2}x}=\frac{1+{\log }_{2}x}{2}[\because \log a^m=m\log a]$
$\Rightarrow {\left({\log }_{2}x\right)}^2-2{\log }_{2}x+1=0$
$\Rightarrow {\log }_{2}x=1$
$\Rightarrow x=2$, which is a prime number.
Ans: B