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If x= z and y=a satisfys the equation 2^(x+y) = 6^y and 3^(x1)= 2^(y+1) then z+a=?
 one year ago
 one year ago
If x= z and y=a satisfys the equation 2^(x+y) = 6^y and 3^(x1)= 2^(y+1) then z+a=?
 one year ago
 one year ago

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open_study1Best ResponseYou've already chosen the best response.1
options are \[\log_{3}6 \] \[\log_{3/2} 6\]
 one year ago

open_study1Best ResponseYou've already chosen the best response.1
@satellite73 @Callisto @experimentX @badi
 one year ago

shivam_bhallaBest ResponseYou've already chosen the best response.0
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
 one year ago

experimentXBest ResponseYou've already chosen the best response.0
\[ 6 = 2^{log_2 6} and \; 3 = 2^{log_2 3}\] do some algebric manipulation
 one year ago

shivam_bhallaBest ResponseYou've already chosen the best response.0
Also log 6 = log(2*3) = log 2 + log 3
 one year ago

open_study1Best ResponseYou've already chosen the best response.1
i am stuck @experimentX @satellite73 @Callisto plz help
 one year ago

experimentXBest ResponseYou've already chosen the best response.0
just some algebraic manipulation ... cancel out the bases .... equate the powers two eqautions ... two unkonwsn ... solve them!!
 one year ago

open_study1Best ResponseYou've already chosen the best response.1
oh......no !!!!! getting an error plzz help
 one year ago

siddhantsharanBest ResponseYou've already chosen the best response.3
Yeah you should get : \[\log_{3/2}6 \]
 one year ago

open_study1Best ResponseYou've already chosen the best response.1
the correct answer is that which @siddhantsharan wrote
 one year ago

experimentXBest ResponseYou've already chosen the best response.0
\[ (x+y) = y\log_2 6 \] \[ (x1)\log_23 = (y+1)\] try solving them
 one year ago

siddhantsharanBest ResponseYou've already chosen the best response.3
Dont solve them. Manupilate to get x+ y.
 one year ago

siddhantsharanBest ResponseYou've already chosen the best response.3
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)
 one year ago

siddhantsharanBest ResponseYou've already chosen the best response.3
From the first you get x log2 = y
 one year ago

siddhantsharanBest ResponseYou've already chosen the best response.3
Then Add the equations And solve for x + y
 one year ago

open_study1Best ResponseYou've already chosen the best response.1
@shivam_bhalla http://openstudy.com/study?signup#/updates/4fc21268e4b0964abc83988e
 one year ago

open_study1Best ResponseYou've already chosen the best response.1
@shivam_bhalla u visited my question.
 one year ago

open_study1Best ResponseYou've already chosen the best response.1
@satellite73 @Mertsj
 one year ago

MertsjBest ResponseYou've already chosen the best response.0
I added the equations like you said but couldn't make any progress from there.
 one year ago

siddhantsharanBest ResponseYou've already chosen the best response.3
Okay. Gimme a minute to type it. :)
 one year ago

siddhantsharanBest ResponseYou've already chosen the best response.3
Assume all logs as base 3 unless written otherwise. \[y = xlog2 \] \[x  1 = ylog2  \log2\] Adding, \[(x + y)  1 = (\log2)(x+y)\]
 one year ago

siddhantsharanBest ResponseYou've already chosen the best response.3
Let x + y be t. t  1 = tlog2 => t = 1 / ( 1  log2)
 one year ago

siddhantsharanBest ResponseYou've already chosen the best response.3
dw:1338210511343:dw
 one year ago

siddhantsharanBest ResponseYou've already chosen the best response.3
@Mertsj Did you understand?
 one year ago

MertsjBest ResponseYou've already chosen the best response.0
I think so except I thought the one equation was \[x1=\log2(y+1)\]
 one year ago

siddhantsharanBest ResponseYou've already chosen the best response.3
Oops. My bad. It is y + 1. Sorry. Steps remain same except answer becomes log(base 3/2) 6
 one year ago

MertsjBest ResponseYou've already chosen the best response.0
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.
 one year ago

siddhantsharanBest ResponseYou've already chosen the best response.3
Gladtoo. Anytime. Your Welcome :)
 one year ago
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