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

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

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Other answers:

\[ 6 = 2^{log_2 6} and \; 3 = 2^{log_2 3}\] 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!! !!!!! 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 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. :)
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)
@Mertsj Did you understand?
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
I think that now I can figure it out if I study it. Thank you so much. I have been working on this for a long time. Don't ask me why because I am just an old lady and it makes no difference at all. Thank so much.
Gladtoo. Anytime. Your Welcome :)

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