If x= z and y=a satisfys the equation 2^(x+y) = 6^y and 3^(x-1)= 2^(y+1) then z+a=?

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

@satellite73 @Callisto @experimentX @badi

Take log on both sides \[\large \log(2^{x+y}) = \log(6^y)\] Applying power rule, we get (x+y)log 2 = y log 6 Similarly apply for the 2nd equation and get two equations and try to solve

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

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

i am stuck @experimentX @satellite73 @Callisto plz help

just some algebraic manipulation ... cancel out the bases .... equate the powers two eqautions ... two unkonwsn ... solve them!!

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.

Take everything on a base 3 by taking log on both sides. \[(x + y)\log_{3}2 = y \log_{3} 6\] And x -1 = log 2(y +1)

From the first you get x log2 = y

Then Add the equations And solve for x + y

@shivam_bhalla http://openstudy.com/study?signup#/updates/4fc21268e4b0964abc83988e

@shivam_bhalla u visited my question.

@satellite73 @Mertsj

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

Okay. Gimme a minute to type it. :)

Assume all logs as base 3 unless written otherwise. \[y = xlog2 \] \[x - 1 = ylog2 - \log2\] Adding, \[(x + y) - 1 = (\log2)(x+y)\]

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

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