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Logarithm simplification

  1. Sep 13, 2010 #1
    1. The problem statement, all variables and given/known data
    Simplify x^(3logx2 - logx5) to find an exact numerical value.


    2. Relevant equations



    3. The attempt at a solution
    3logx2=logx2^3 or logx8,
    (logx8 - logx5)=logx8/5
    the inverse would be x^y=8/5 (y is unknown)
    therefore logx8/5=y and x^(logx8/5)=x^y=8/5
    and the inverse of that would be logx8/5=logx8/5

    (logx8/5)-(logx8/5)=0
    rewriting it you get logx(8/5)/(8/5) or just logx1=0
    once again the inverse would be 1=x^0 which tells me that x can be any given number.
    Is this correct?
     
  2. jcsd
  3. Sep 13, 2010 #2

    LCKurtz

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    Science Advisor
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    Gold Member

    Re: Logarithms

    Your question asked you to just simplify the expression. You have shown that the expression given reduces to 8/5, which is correct, and you could stop there. And you are also correct that it doesn't matter what x is, although you would want x > 0.
     
  4. Sep 14, 2010 #3
    Re: Logarithms

    Sorry for the late response, for some reason I couldn't login to my account yesterday.

    Your right, I was trying to find an exact numerical value for x.

    Shouldn't it be x>1, since 1^y always equals 1.
     
    Last edited: Sep 14, 2010
  5. Sep 14, 2010 #4

    Mark44

    Staff: Mentor

    Re: Logarithms

    Right, the base shouldn't be 1 either. So the conditions are x > 0 and x [itex]\neq[/itex] 1.
     
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