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open_study1

  • 2 years ago

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

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  1. open_study1
    • 2 years ago
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    options are \[\log_{3}6 \] \[\log_{3/2} 6\]

  2. open_study1
    • 2 years ago
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    @satellite73 @Callisto @experimentX @badi

  3. shivam_bhalla
    • 2 years ago
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    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

  4. experimentX
    • 2 years ago
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    \[ 6 = 2^{log_2 6} and \; 3 = 2^{log_2 3}\] do some algebric manipulation

  5. shivam_bhalla
    • 2 years ago
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    Also log 6 = log(2*3) = log 2 + log 3

  6. open_study1
    • 2 years ago
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    i am stuck @experimentX @satellite73 @Callisto plz help

  7. experimentX
    • 2 years ago
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    just some algebraic manipulation ... cancel out the bases .... equate the powers two eqautions ... two unkonwsn ... solve them!!

  8. open_study1
    • 2 years ago
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    oh......no !!!!! getting an error plzz help

  9. siddhantsharan
    • 2 years ago
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    Yeah you should get : \[\log_{3/2}6 \]

  10. open_study1
    • 2 years ago
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    the correct answer is that which @siddhantsharan wrote

  11. experimentX
    • 2 years ago
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    \[ (x+y) = y\log_2 6 \] \[ (x-1)\log_23 = (y+1)\] try solving them

  12. siddhantsharan
    • 2 years ago
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    Dont solve them. Manupilate to get x+ y.

  13. siddhantsharan
    • 2 years ago
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    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)

  14. siddhantsharan
    • 2 years ago
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    From the first you get x log2 = y

  15. siddhantsharan
    • 2 years ago
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    Then Add the equations And solve for x + y

  16. open_study1
    • 2 years ago
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    @shivam_bhalla http://openstudy.com/study?signup#/updates/4fc21268e4b0964abc83988e

  17. open_study1
    • 2 years ago
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    @shivam_bhalla u visited my question.

  18. open_study1
    • 2 years ago
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    @satellite73 @Mertsj

  19. Mertsj
    • 2 years ago
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    I added the equations like you said but couldn't make any progress from there.

  20. siddhantsharan
    • 2 years ago
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    Okay. Gimme a minute to type it. :)

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

  22. siddhantsharan
    • 2 years ago
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    Let x + y be t. t - 1 = tlog2 => t = 1 / ( 1 - log2)

  23. siddhantsharan
    • 2 years ago
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    |dw:1338210511343:dw|

  24. siddhantsharan
    • 2 years ago
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    @Mertsj Did you understand?

  25. Mertsj
    • 2 years ago
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    I think so except I thought the one equation was \[x-1=\log2(y+1)\]

  26. siddhantsharan
    • 2 years ago
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    Oops. My bad. It is y + 1. Sorry. Steps remain same except answer becomes log(base 3/2) 6

  27. Mertsj
    • 2 years ago
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    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.

  28. siddhantsharan
    • 2 years ago
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    Gladtoo. Anytime. Your Welcome :)

  29. Mertsj
    • 2 years ago
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    |dw:1338234189320:dw|

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