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If (log2x)/4=(log2y)/6=(log2z)/(3k) and...

If `(log_2x)/4=(log_2y)/6=(log_2z)/(3k)` and `x^3 y^2 z=1`, then k is equal to

A

`-8`

B

`-4`

C

0

D

4

Text Solution

Verified by Experts

The correct Answer is:
A
`(log_(2)x)/(4)=(log_(2)y)/(6)=(log_(2)z)/(3k)`
`=(3log_(2)x + 2 log_(2)y + log_(2)z)/(12+12+3k)`
`(log_(2)(x^(3)y^(2)x))/(24+3k)`
`=(0)/(24+3k)`
`rArr 24 + 3k = 0 rArr k k=-8`
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