Question

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

Clearly, $x>0,$ $\sqrt{x}>0$
$\Rightarrow x>0\cdots \left(1\right)$
and ${\log }_{2}x\ge 0$
$\Rightarrow x\ge 1\cdots \left(2\right)$
So, from $\left(1\right)$ and $\left(2\right),$ $x\ge 1$

$\sqrt{{\log }_{2}x}-0.5={\log }_2\sqrt{x}$
$\Rightarrow \sqrt{{\log }_{2}x}=0.5+0.5{\log }_{2}x$
$\Rightarrow {\log }_{2}x=\frac{1}{4}(1+2{\log }_{2}x+({\log }_{2}x{)}^2)$
$\Rightarrow 4{\log }_{2}x=1+2{\log }_{2}x+({\log }_{2}x{)}^2$
$\Rightarrow ({\log }_{2}x-1{)}^2=0$
$\Rightarrow {\log }_{2}x=1$
$\Rightarrow x=2$