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Math Help - Logarithms

  1. #1
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    Logarithms

    How would I go about solving this problem?


     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?


     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  2 = e^{5x} ..

    so here are some reviews..

    \ln e = 1

    \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,

    2 = e^{5x}


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

    =>  .6931 = 5x ln e

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

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

    =>  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 \ln a
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  5. #5
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    ok. And thanks
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