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options are \[\log_{3}6 \] \[\log_{3/2} 6\]

\[ 6 = 2^{log_2 6} and \; 3 = 2^{log_2 3}\]
do some algebric manipulation

Also log 6 = log(2*3) = log 2 + log 3

oh......no !!!!! getting an error plzz help

Yeah you should get : \[\log_{3/2}6 \]

the correct answer is that which @siddhantsharan wrote

\[ (x+y) = y\log_2 6 \]
\[ (x-1)\log_23 = (y+1)\]
try solving them

Dont solve them. Manupilate to get x+ y.

From the first you get x log2 = y

Then Add the equations And solve for x + y

@shivam_bhalla u visited my question.

I added the equations like you said but couldn't make any progress from there.

Okay. Gimme a minute to type it. :)

Let x + y be t.
t - 1 = tlog2 => t = 1 / ( 1 - log2)

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I think so except I thought the one equation was
\[x-1=\log2(y+1)\]

Oops. My bad.
It is y + 1. Sorry.
Steps remain same except answer becomes log(base 3/2) 6

Gladtoo. Anytime. Your Welcome :)

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