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OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

@satellite73 @Callisto @experimentX @badi

OpenStudy (anonymous):

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

OpenStudy (experimentx):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

i am stuck @experimentX @satellite73 @Callisto plz help

OpenStudy (experimentx):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

the correct answer is that which @siddhantsharan wrote

OpenStudy (experimentx):

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

OpenStudy (anonymous):

Dont solve them. Manupilate to get x+ y.

OpenStudy (anonymous):

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)

OpenStudy (anonymous):

From the first you get x log2 = y

OpenStudy (anonymous):

Then Add the equations And solve for x + y

OpenStudy (anonymous):

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

OpenStudy (anonymous):

@shivam_bhalla u visited my question.

OpenStudy (anonymous):

@satellite73 @Mertsj

OpenStudy (mertsj):

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

OpenStudy (anonymous):

Okay. Gimme a minute to type it. :)

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

|dw:1338210511343:dw|

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