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Math Help - Impossible(for me) Logarithm Proof Please Help!

  1. #1
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    Impossible(for me) Logarithm Proof Please Help!

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    Any hints or anything to steer me in the right direction would be awesome!
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  2. #2
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    By law of indices



    nth root of (x to the power of m) becomes x to the power of m/n

    gives log_b x^(m/n)
    Then by

    Power Law of Logarithms:
    log a^x = x log a

    we get m/nlog_b x

    sorry about fonts!
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  3. #3
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    I have to prove the law, I can not rely on it. Also where does the b^t play in?
    We apparently have to use it.
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  4. #4
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    I'm pretty much stuck with jaydee323 here.
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  5. #5
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    Well, we know that \sqrt[n]{a^m} = a^{\frac{m}{n}}.

    So we just have to prove that: \log_{b} a^{\frac{m}{n}} = \frac{m}{n} \log_{b} a

    Let: {\color{red}u = \log_{b}a} \ \iff \ b^u = a

    Raise both sides to the m/n: \left(b^u\right)^{\displaystyle m/n} = a^{\displaystyle m/n} \ \iff \ b^{\displaystyle mu/n} = a^{\displaystyle m/n} \ \iff \ \log_{b} \left(a^{\displaystyle m/n}\right) = \frac{m{\color{red}u}}{n} = \hdots
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