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Does anyone know how to solve this log question ?

  1. Jan 21, 2009 #1
    show how
     
    Last edited: Jan 21, 2009
  2. jcsd
  3. Jan 21, 2009 #2

    rock.freak667

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    You'll need these formulas

    [tex]nlog_a b= log_a b^n[/tex]

    [tex] log_a b= log_a c \Rightarrow b=c[/tex]

    [tex]log_a x + log_a y = log_a xy[/tex]
     
  4. Jan 21, 2009 #3

    rock.freak667

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    you mean you'd get log(x-y/3)= 1/2log(xy)?

    Use first formula on 1/2log(xy) and then apply the second formula.
     
  5. Jan 21, 2009 #4

    rock.freak667

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    Yes.

    [tex]\frac{x-y}{3} =(xy)^{\frac{1}{2}}[/tex]

    Square both sides now.
     
  6. Jan 21, 2009 #5
    Thank-you very much. Question solved.
     
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