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open_study1
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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=?
 2 years ago
 2 years ago
open_study1 Group Title
If x= z and y=a satisfys the equation 2^(x+y) = 6^y and 3^(x1)= 2^(y+1) then z+a=?
 2 years ago
 2 years ago

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open_study1 Group TitleBest ResponseYou've already chosen the best response.1
options are \[\log_{3}6 \] \[\log_{3/2} 6\]
 2 years ago

open_study1 Group TitleBest ResponseYou've already chosen the best response.1
@satellite73 @Callisto @experimentX @badi
 2 years ago

shivam_bhalla Group TitleBest 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
 2 years ago

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

shivam_bhalla Group TitleBest ResponseYou've already chosen the best response.0
Also log 6 = log(2*3) = log 2 + log 3
 2 years ago

open_study1 Group TitleBest ResponseYou've already chosen the best response.1
i am stuck @experimentX @satellite73 @Callisto plz help
 2 years ago

experimentX Group TitleBest 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!!
 2 years ago

open_study1 Group TitleBest ResponseYou've already chosen the best response.1
oh......no !!!!! getting an error plzz help
 2 years ago

siddhantsharan Group TitleBest ResponseYou've already chosen the best response.3
Yeah you should get : \[\log_{3/2}6 \]
 2 years ago

open_study1 Group TitleBest ResponseYou've already chosen the best response.1
the correct answer is that which @siddhantsharan wrote
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.0
\[ (x+y) = y\log_2 6 \] \[ (x1)\log_23 = (y+1)\] try solving them
 2 years ago

siddhantsharan Group TitleBest ResponseYou've already chosen the best response.3
Dont solve them. Manupilate to get x+ y.
 2 years ago

siddhantsharan Group TitleBest 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)
 2 years ago

siddhantsharan Group TitleBest ResponseYou've already chosen the best response.3
From the first you get x log2 = y
 2 years ago

siddhantsharan Group TitleBest ResponseYou've already chosen the best response.3
Then Add the equations And solve for x + y
 2 years ago

open_study1 Group TitleBest ResponseYou've already chosen the best response.1
@shivam_bhalla http://openstudy.com/study?signup#/updates/4fc21268e4b0964abc83988e
 2 years ago

open_study1 Group TitleBest ResponseYou've already chosen the best response.1
@shivam_bhalla u visited my question.
 2 years ago

open_study1 Group TitleBest ResponseYou've already chosen the best response.1
@satellite73 @Mertsj
 2 years ago

Mertsj Group TitleBest ResponseYou've already chosen the best response.0
I added the equations like you said but couldn't make any progress from there.
 2 years ago

siddhantsharan Group TitleBest ResponseYou've already chosen the best response.3
Okay. Gimme a minute to type it. :)
 2 years ago

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

siddhantsharan Group TitleBest ResponseYou've already chosen the best response.3
Let x + y be t. t  1 = tlog2 => t = 1 / ( 1  log2)
 2 years ago

siddhantsharan Group TitleBest ResponseYou've already chosen the best response.3
dw:1338210511343:dw
 2 years ago

siddhantsharan Group TitleBest ResponseYou've already chosen the best response.3
@Mertsj Did you understand?
 2 years ago

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

siddhantsharan Group TitleBest 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
 2 years ago

Mertsj Group TitleBest 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.
 2 years ago

siddhantsharan Group TitleBest ResponseYou've already chosen the best response.3
Gladtoo. Anytime. Your Welcome :)
 2 years ago

Mertsj Group TitleBest ResponseYou've already chosen the best response.0
dw:1338234189320:dw
 2 years ago
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