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Thread: Logarithm

  1. #1
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    Logarithm

    Hi,

    Are there alternative ways of solving this equation? Could logarithms be used?

    $\displaystyle a^{60} = 2.044 \implies a = 2.044^{1/60} = 1.012$
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  2. #2
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    Quote Originally Posted by Hellbent View Post
    Hi,

    Are there alternative ways of solving this equation? Could logarithms be used?

    $\displaystyle a^{60} = 2.044 \implies a = 2.044^{1/60} = 1.012$
    logs could be used ... but the way you did it more direct.
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  3. #3
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    I need assistance with the logs part, please.
    Tried using logs prior to posting question, but the answers are inconsistent.
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  4. #4
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    Quote Originally Posted by Hellbent View Post
    Hi,

    Are there alternative ways of solving this equation? Could logarithms be used?

    $\displaystyle a^{60} = 2.044 \implies a = 2.044^{1/60} = 1.012$
    $\displaystyle 60\log{a} = \log{2.044}$

    $\displaystyle \log{a} = \frac{log{2.044}}{60}
    \RightArrow
    = .00517
    $

    so $\displaystyle 10^{(.00517)} = 1.01199$ or $\displaystyle 1.012$

    as mentioned your first method is better..
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  5. #5
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    $\displaystyle a^{60} = k$ , where $\displaystyle k = 2.044$

    $\displaystyle 60\log{a} = \log{k}
    $

    $\displaystyle \displaystyle \log{a} = \frac{\log{k}}{60}$

    $\displaystyle \displaystyle a = e^{\frac{\log{k}}{60}}$

    $\displaystyle a \approx 1.012$

    note that $\displaystyle \displaystyle a = e^{\frac{\log{k}}{60}} = \left(e^{\log{k}}\right)^{\frac{1}{60}} = k^{\frac{1}{60}} = 2.044^{\frac{1}{60}}$
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