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

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

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

shivam_bhalla
 2 years ago
Best ResponseYou've already chosen the best response.0Take 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

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

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

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

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.0just some algebraic manipulation ... cancel out the bases .... equate the powers two eqautions ... two unkonwsn ... solve them!!

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

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

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

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

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

siddhantsharan
 2 years ago
Best ResponseYou've already chosen the best response.3Take 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)

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

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

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

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

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

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

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

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

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

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

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

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

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

Mertsj
 2 years ago
Best ResponseYou've already chosen the best response.0I 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.

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