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Thread: log problem

  1. #1
    Member Veronica1999's Avatar
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    log problem

    Please find attached problem and my work.
    Attached Thumbnails Attached Thumbnails log problem-log.pdf  
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by Veronica1999 View Post
    Please find attached problem and my work.
    The integers with rational logs to the base 8 are all of the form $\displaystyle 2^k$ for some natural number $\displaystyle $$k$.

    CB
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  3. #3
    Member Veronica1999's Avatar
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    How do I prove that log(base8)6 has an irrational value?
    I can split this into log(base8)2 + log(base8)3 = 1/3 + ?
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  4. #4
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    Quote Originally Posted by Veronica1999 View Post
    How do I prove that log(base8)6 has an irrational value?
    I can split this into log(base8)2 + log(base8)3 = 1/3 + ?
    Proof by contradiction. Assume it is rational.

    $\displaystyle \log_{8} 6 = \frac{a}{b}$

    $\displaystyle 8^{\frac{a}{b}} = 6$

    $\displaystyle 8^a = 6^b$

    $\displaystyle 2^{3a} = 2^b*3^b$

    $\displaystyle 2^{3a-b} = 3^b$

    From unique factorization, this is impossible - a contradiction.

    QED.
    Last edited by elemental; Dec 1st 2010 at 01:28 PM. Reason: formatting
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