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Thread: Logarithms

  1. #1
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    Logarithms

    How would I go about solving this problem?


    $\displaystyle 2 = e^5x $ (5x being the exponent)

    The whole natural logarithms, antilogarithms, and Euler's number stuff isn't really clicking in my head.
    If you explained it step by step that would help a lot.
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  2. #2
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by Hypertension View Post
    How would I go about solving this problem?


    $\displaystyle 2 = e^5x $ (5x being the exponent)

    The whole natural logarithms, antilogarithms, and Euler's number stuff isn't really clicking in my head.
    If you explained it step by step that would help a lot.
    you mean $\displaystyle 2 = e^{5x} $..

    so here are some reviews..

    $\displaystyle \ln e = 1$

    $\displaystyle \ln a^{b} = b \ln a$

    this must be sufficient..

    first, take the natural logarithms of both sides..
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  3. #3
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    so,

    $\displaystyle 2 = e^{5x}$


    => $\displaystyle (ln)2 = (ln)e^{5x}$

    => $\displaystyle .6931 = 5x ln e $

    => $\displaystyle .6931 = 5x $ (since $\displaystyle ln e = 1 $, right?)

    => $\displaystyle \frac{.6931}{5} = \frac{5x}{5} $

    => $\displaystyle x = .1386 $



    ?
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  4. #4
    MHF Contributor kalagota's Avatar
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    bravo!!


    anyways, put \ before ln i.e
    Code:
    \ln a
    will look like $\displaystyle \ln a$
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  5. #5
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    ok. And thanks
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