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Thread: logs again

  1. #1
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    logs again

    simplify log(base4/9)(27/16) without a calculator

    i've tried everything i could think of without a calculator, so here i am.
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  2. #2
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    Quote Originally Posted by purebladeknight View Post
    simplify log(base4/9)(27/16) without a calculator

    i've tried everything i could think of without a calculator, so here i am.
    $\displaystyle \frac{27}{16} = \frac{3}{4}\cdot\frac{9}{4}$

    $\displaystyle = \frac{3}{4}\left(\frac{4}{9}\right)^{-1}$.


    So $\displaystyle \log_{\frac{4}{9}}\frac{27}{16} = \log_{\frac{4}{9}}\left[\frac{3}{4}\left(\frac{4}{9}\right)^{-1}\right]$

    $\displaystyle = \log_{\frac{4}{9}}\frac{3}{4} + \log_{\frac{4}{9}}\left(\frac{4}{9}\right)^{-1}$

    $\displaystyle = \log_{\frac{4}{9}}\frac{3}{4} - 1$

    $\displaystyle = \log_{\frac{4}{9}}{3} - \log_{\frac{4}{9}}{4} - 1$.
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  3. #3
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    Quote Originally Posted by Prove It View Post
    $\displaystyle \frac{27}{16} = \frac{3}{4}\cdot\frac{9}{4}$

    $\displaystyle = \frac{3}{4}\left(\frac{4}{9}\right)^{-1}$.


    So $\displaystyle \log_{\frac{4}{9}}\frac{27}{16} = \log_{\frac{4}{9}}\left[\frac{3}{4}\left(\frac{4}{9}\right)^{-1}\right]$

    $\displaystyle = \log_{\frac{4}{9}}\frac{3}{4} + \log_{\frac{4}{9}}\left(\frac{4}{9}\right)^{-1}$

    $\displaystyle = \log_{\frac{4}{9}}\frac{3}{4} - 1$

    $\displaystyle = \log_{\frac{4}{9}}{3} - \log_{\frac{4}{9}}{4} - 1$.
    silly me, i was trapped in the thought of solving it! thanks for the enlightenment on my silly misunderstanding
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  4. #4
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    Hello purebladeknight
    Quote Originally Posted by purebladeknight View Post
    simplify log(base4/9)(27/16) without a calculator

    i've tried everything i could think of without a calculator, so here i am.
    Here's an alternative approach.

    Let $\displaystyle x = \log_{\frac49}\left(\frac{27}{16}\right)$

    Then $\displaystyle \left(\frac49\right)^x = \frac{27}{16}$

    $\displaystyle \Rightarrow \left(\left(\frac23\right)^2\right)^x = \frac{3^3}{2^4}$

    $\displaystyle \Rightarrow \left(\frac23\right)^{2x}=\frac12\left(\frac23\rig ht)^{-3}$

    $\displaystyle \Rightarrow 2x = \log_{\frac23}\left(\frac12\right)-3\log_{\frac23}\left(\frac23\right)$
    $\displaystyle =-\log_{\frac23}(2)-3$
    $\displaystyle \Rightarrow x = -\tfrac12(\log_{\frac23}(2)+3)$

    ...but whether you think this answer is any simpler is a matter for debate!

    If you want to get rid of fractional bases, you can express this in terms of $\displaystyle \log_2$ using the formula
    $\displaystyle \log_{\frac23}(2) = \frac{\log_22}{\log_2(\tfrac23)}$
    $\displaystyle =\frac{1}{1-\log_23}$
    which then gives:
    $\displaystyle x = \frac{4-3\log_23}{2(\log_23-1)}$
    but I don't think you can rid of logs altogether!

    Grandad
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