Question

If $x{y}^2=4$ and ${\log }_3\left({\log }_{2}x\right)+{\log }_{1/3}\left({\log }_{1/2}y\right)=1$, then $x$ is equal to

• A. $4$
• B. $8$
• C. $16$
• D. $64$

Answer 1

The correct option is D $64$

$x{y}^2=4$ ...(1)
${\log }_3\left({\log }_{2}x\right)+{\log }_{\frac{1}{3}}\left({\log }_{\frac{1}{2}}y\right)=1\Rightarrow {\log }_3\left({\log }_{2}x\right)-{\log }_3\left({\log }_{\frac{1}{2}}y\right)=1\Rightarrow {\log }_3\left(\frac{{\log }_{2}x}{{\log }_{\frac{1}{2}}y}\right)=1\Rightarrow \frac{{\log }_{2}x}{-{\log }_{2}y}=3\Rightarrow {\log }_{2}x={\log }_2{y}^{-3}$
$\Rightarrow x{y}^3=1$ ...(2)
From (1) & (2)
$x=64\&y=\frac{1}{4}$
Ans: D