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Logarithm with different base

  1. Aug 13, 2009 #1
    1. The problem statement, all variables and given/known data
    Hi everyone

    I need help for this problem :

    If [tex]2*\log_2 (x-2y)=\log_3 (xy)[/tex] , find [tex]\frac{x}{y}[/tex]


    2. Relevant equations
    [tex]\log_bx = \frac{\log_ax}{\log_ab}[/tex]


    3. The attempt at a solution
    [tex]2*\log_2 (x-2y)=\log_3 (xy)[/tex]

    [tex]\log_2 (x-2y)^2=\log_3 (xy)[/tex]

    [tex]\frac{\log_2 (x-2y)^2}{\log_2 2}=\frac{\log_2 (xy)}{\log_2 3}[/tex]

    [tex]\log_2 (x-2y)^2 * \log_2 3 = \log_2 (xy)[/tex]

    Then, I stuck .....

    Thx :)
     
  2. jcsd
  3. Aug 14, 2009 #2

    rock.freak667

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    Homework Helper

    Are you sure you wrote the question down correctly? I ask because log23 is not an integer and you be basically computing (x-2y)k=xy where k is not an integer (so you'd not be able to find x/y)
     
  4. Aug 14, 2009 #3
    Hi rock.freak667

    At least that's the whole question that was given to me by my friend. Your post assure me that this question can't be solved.

    Thx a lot for pointing out that [tex]\log_2 3[/tex] is not an integer :smile:
     
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