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Q.

If xy2=4 and log3⁡log2⁡x+log1/3⁡log1/2⁡y=1,  then x=

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a

4

b

8

c

16

d

64

answer is D.

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Detailed Solution

log3⁡log2⁡x+log1/3⁡log1/2⁡y=1 or  log3⁡log2⁡x−log3⁡log1/2⁡y=1 or  log3⁡log2⁡4/y2−log3⁡log1/2⁡y=1 or  log2⁡4/y2=3log1/2⁡y or  log2⁡4/y2=−3log2⁡y or  log2⁡4/y2+log2⁡y3=0 or  4y=1 or  y=1/4⇒ x=64
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