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Math Help - Simple Logarithms Question

  1. #1
    Newbie evanator's Avatar
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    Question Simple Logarithms Question

    Hi all,

    I have returned to maths after a break of five years. Although I was quite good at it in school, I studied a non-technical subject at university and I have lost nearly all my mathematical ability.

    I encountered this problem while attempting to revise logarithms:

    Use logarithms to solve the following equation:

    10^x = e^{2x-1}

    I attempted to solve it as follows:

    x \log 10 = 2 x \log e - x \log e

    \log e = 2 x \log e - x \log e

    \log e = x \log ( \frac {e^2}{10} )

    x = \frac {\log e}{\log \frac {e^2}{10}}

    x = \frac {0.43429}{-0.13141}

    x = -13.82649

    This is not the correct answer. I would be grateful if somebody told me where I am going wrong.

    Regards,

    Evanator
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  2. #2
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    when you log both sides it does not cancel out the e. the only thing that can cancel out an e is a ln.
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  3. #3
    A riddle wrapped in an enigma
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    Hi evanator,

    Use natural logs.

    10^x=e^{2x-1}

    \ln 10^x=\ln e^{2x-1}

    x \ln 10=2x-1

    2.302585093 x = 2x-1



    Now, can you finish up?
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  4. #4
    Newbie evanator's Avatar
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    Smile

    0.302585093x = -1

    x = - \frac {1}{0.302585093}

    x = -3.304855


    Thank you both for your help, which I sincerely appreciate. I hope my brain kicks in to gear again soon.

    Later,

    evanator
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  5. #5
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    e^(i*pi)'s Avatar
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    Quote Originally Posted by evanator View Post
    0.302585093x = -1

    x = - \frac {1}{0.302585093}

    x = -3.304855


    Thank you both for your help, which I sincerely appreciate. I hope my brain kicks in to gear again soon.

    Later,

    evanator
    Another way:

    10^x = e^{2x-1}

    \ln (10^x) = \ln(e^{2x-1})

    x \ln (10) = 2x-1<br />

    2x - x \ln (10) = 1

    x(2-\ln(10)) = 1

    x = \frac{1}{2-\ln(10)}
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