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A Log Questions for ya

  1. Dec 10, 2005 #1
  2. jcsd
  3. Dec 11, 2005 #2

    Tide

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    Not too tricky - it's just the Golden Ratio.
     
  4. Dec 11, 2005 #3

    JasonRox

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  5. Dec 12, 2005 #4
    Yar, still can't get it. Answer anybody :smile:
     
  6. Dec 13, 2005 #5

    Tide

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    The answer is in #2 above.
     
  7. Dec 10, 2010 #6

    Avodyne

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    The problem is to find [itex]a/b[/itex] given
    [tex]\log_9 a = \log_{12}b = \log_{16}(a+b)[/tex]

    Let [tex]x = \log_9 a = \log_{12}b = \log_{16}(a+b)[/tex]. Then

    [tex]a=9^x,~~~b=12^x,~~~a+b=16^x[/tex]

    Now, compare [itex]a(a+b)[/itex] with [itex]b^2[/itex].
     
  8. Dec 10, 2010 #7

    berkeman

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    We do not give out answers here at the PF. Show some effort, or your thread will be deleted.
     
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