If $x y^{2} = 4$ and $l o g_{3} (l o g_{2} x) + l o g_{1 / 3} (l o g_{1 / 2} y) = 1$ , then x equals

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Solution

The correct option is D
64

$l o g_{3} (l o g_{2} x) + l o g_{1 / 3} (l o g_{1 / 2} y) = 1 l o g_{3} l o g_{2} x - l o g_{3} l o g_{\frac{1}{2}} y = 1 l o g_{3} \frac{l o g_{2} x}{l o g_{\frac{1}{2}} y} = 1 \Rightarrow \frac{l o g_{2} x}{l o g_{\frac{1}{2}} y} = 3 \Rightarrow \frac{l o g_{2} x}{- l o g_{2} y} = 3 l o g_{2} x = - 3 l o g_{2} y \to (1) x y^{2} = 4 l o g_{2} x + 2 l o g_{2} y = 2 l o g_{2} 2 l o g_{2} y = \frac{2 - l o g_{2} x}{2} \to (2)$
$(2) i n (1) l o g_{2} x = - 3 (\frac{2 - 2 l o g_{2} x}{2}) 2 l o g_{2} x = - 6 + 3 l o g_{2} x 6 = l o g_{2} x x = 2^{6} x = 64$

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