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Steps on how to simplify log5/log125 to 1/3

  1. Feb 24, 2005 #1
    could someone show me the steps on how to simplify log5/log125 to 1/3. I cant do it :grumpy:
     
  2. jcsd
  3. Feb 24, 2005 #2
    Which base?
     
  4. Feb 24, 2005 #3
    10, sorry for not mentioning that.
     
  5. Feb 24, 2005 #4

    dextercioby

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    It doesn't matter the base,as long it is the same...:wink:

    Daniel.
     
  6. Feb 24, 2005 #5
    yes, so how could I simplify it?
     
  7. Feb 24, 2005 #6

    dextercioby

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    [itex] 125=5^{3} [/itex] and use one definitory property of the logarithm...

    Daniel.
     
  8. Feb 24, 2005 #7
    hmmm....that comes to log5 to the power -2?
     
  9. Feb 24, 2005 #8
    Why don't just: log5 125 = 3 and log5 5 = 1 and then just replace log 5 / log 125 = 1 / 3 ??
     
  10. Feb 24, 2005 #9

    dextercioby

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    No,remember that in general:
    [tex] \frac{\log a}{\log b}\neq \log\frac{a}{b} [/tex]

    So pay attention to what u do.
    Daniel.
     
  11. Feb 24, 2005 #10
    log(125) = log(53) = 3log(5)
     
  12. Feb 24, 2005 #11
    oh right. My bad. so log5/log125 = log5/log5^3 = log5/3log5, the log5's cancel, leaving 1/3. Thanx
     
  13. Feb 24, 2005 #12
    And then cancel out the log(5)s, leaving 1/3?
     
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