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Math Help - Proof of Logarithms

  1. #1
    Junior Member masoug's Avatar
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    Question Proof of Logarithms

    Hi,
    I was looking at a math book when I stumbled across a problem:
    Prove the following:
    log x < x given x > 0
    I am a little puzzled. How would I prove it? Using proof by counterexample?

    Thanks!
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  2. #2
    MHF Contributor
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    You can prove this using derivatives...
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  3. #3
    Junior Member masoug's Avatar
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    Okay... Is it possible without derivatives?
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  4. #4
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    Why would you want a different method?

    First note that \displaystyle \ln{x} < 0 for all \displaystyle x \in (0,1), while \displaystyle x > 0 for all \displaystyle x > 0. So \displaystyle \ln{x} < x for all \displaystyle x \in (0, 1).

    Then note that \displaystyle \frac{d}{dx}(\ln{x}) = \frac{1}{x} and \displaystyle \frac{d}{dx}(x) = 1.

    For \displaystyle x \geq 1, we have \displaystyle \frac{1}{x} < 1

    So that means \displaystyle \ln{x} grows slower than \displaystyle x does, so can never catch up.

    Therefore \displaystyle \ln{x} < x for all \displaystyle x > 0.
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  5. #5
    Junior Member masoug's Avatar
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    Okay, that makes sense... So we can use growth rates of functions to determine which is the largest in the end.
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  6. #6
    Math Engineering Student
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    Use a contradiction: suppose not, that means \ln x\ge x, that would imply x\ge e^x, but this is obviously false since x>0.
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