anonymous
  • 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=?
Mathematics
schrodinger
  • schrodinger
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anonymous
  • anonymous
options are \[\log_{3}6 \] \[\log_{3/2} 6\]
anonymous
  • anonymous
@satellite73 @Callisto @experimentX @badi
anonymous
  • 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

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experimentX
  • experimentX
\[ 6 = 2^{log_2 6} and \; 3 = 2^{log_2 3}\] do some algebric manipulation
anonymous
  • anonymous
Also log 6 = log(2*3) = log 2 + log 3
anonymous
  • anonymous
i am stuck @experimentX @satellite73 @Callisto plz help
experimentX
  • experimentX
just some algebraic manipulation ... cancel out the bases .... equate the powers two eqautions ... two unkonwsn ... solve them!!
anonymous
  • anonymous
oh......no !!!!! getting an error plzz help
anonymous
  • anonymous
Yeah you should get : \[\log_{3/2}6 \]
anonymous
  • anonymous
the correct answer is that which @siddhantsharan wrote
experimentX
  • experimentX
\[ (x+y) = y\log_2 6 \] \[ (x-1)\log_23 = (y+1)\] try solving them
anonymous
  • anonymous
Dont solve them. Manupilate to get x+ y.
anonymous
  • 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)
anonymous
  • anonymous
From the first you get x log2 = y
anonymous
  • anonymous
Then Add the equations And solve for x + y
anonymous
  • anonymous
@shivam_bhalla u visited my question.
anonymous
  • anonymous
@satellite73 @Mertsj
Mertsj
  • Mertsj
I added the equations like you said but couldn't make any progress from there.
anonymous
  • anonymous
Okay. Gimme a minute to type it. :)
anonymous
  • anonymous
Assume all logs as base 3 unless written otherwise. \[y = xlog2 \] \[x - 1 = ylog2 - \log2\] Adding, \[(x + y) - 1 = (\log2)(x+y)\]
anonymous
  • anonymous
Let x + y be t. t - 1 = tlog2 => t = 1 / ( 1 - log2)
anonymous
  • anonymous
|dw:1338210511343:dw|
anonymous
  • anonymous
@Mertsj Did you understand?
Mertsj
  • Mertsj
I think so except I thought the one equation was \[x-1=\log2(y+1)\]
anonymous
  • anonymous
Oops. My bad. It is y + 1. Sorry. Steps remain same except answer becomes log(base 3/2) 6
Mertsj
  • Mertsj
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.
anonymous
  • anonymous
Gladtoo. Anytime. Your Welcome :)
Mertsj
  • Mertsj
|dw:1338234189320:dw|

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