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

4

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8

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16

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64

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Solution

The correct option is D $64$
$x y^{2} = 4$ ...(1)
${log}_{3} ({log}_{2} x) + {log}_{\frac{1}{3}} ({log}_{\frac{1}{2}} y) = 1 \Rightarrow {log}_{3} ({log}_{2} x) - {log}_{3} ({log}_{\frac{1}{2}} y) = 1 \Rightarrow {log}_{3} ⎛ ⎜ ⎝ \frac{{log}_{2} x}{{log}_{\frac{1}{2}} y} ⎞ ⎟ ⎠ = 1 \Rightarrow \frac{{log}_{2} x}{- {log}_{2} y} = 3 \Rightarrow {log}_{2} x = {log}_{2} y^{- 3}$
$\Rightarrow x y^{3} = 1$ ...(2)
From (1) & (2)
$x = 64 & y = \frac{1}{4}$
Ans: D

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