Question

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

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

Answer 1

Answer: (B)

prime number

Given, $\sqrt{{\log }_{2}x}=0.5+{\log }_2\sqrt{x}$, where $x>0$
and ${\log }_{2}x>0$

$\Rightarrow x>1$
We have ${\log }_2\sqrt{x}=\frac{1}{2}lo{g}_{2}x$
$\Rightarrow \sqrt{{\log }_{2}x}=\frac{1}{2}+\frac{{\log }_{2}x}{2}$
On squaring, we get
${\log }_{2}x=\frac{1+2{\log }_{2}x+({\log }_{2}x{)}^2}{4}$
$\Rightarrow 4{\log }_{2}x=1+2{\log }_{2}x+({\log }_{2}x{)}^2$
$\Rightarrow {\left({\log }_{2}x\right)}^2-2{\log }_{2}x+1=0\Rightarrow {\log }_{2}x=1$

$\Rightarrow x=2$
Hence, $x$ is a prime number
Hence, option 'B' is correct.