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Math Help - Solving Logarithmic Fractions

  1. #1
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    Lightbulb Solving Logarithmic Fractions

    I came upon this question to do with Logs and Fractions which I'm not too sure about. {3log_{4} x} + \frac{\1}{2} = {5log_{8} x}

    I changed the bases to end up with \frac{\3lnx}{ln4} - \frac{\5lnx}{ln8} = \frac{\1}{2}

    But forgot the rules about dividing logarithmic functions. Could somebody point me in the right direction please?

    P.S. The question asks you to solve for x
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  2. #2
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    Re: Solving Logarithmic Fractions

    Take ln x out as a common factor from the left hand side. Divide the remaining term over then take both sides to the power of e.
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  3. #3
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    Re: Solving Logarithmic Fractions

    How do I take ln x out as a common factor? Since the its in the numerator. I don't quite understand that part
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  4. #4
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    Re: Solving Logarithmic Fractions

    Alternately. you could use the identity:

    \log_{b^n}a =\frac{\log_b(a)}{n} to write the equation as:

    3\log_{2^2}(x)+\frac{1}{2}=5\log_{2^3}(x)

    3\frac{\log_2(x)}{2}+\frac{1}{2}=5\frac{\log_2(x)}  {3}

    Now solve for \log_2(x) and convert from logarithmic to exponential form...
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  5. #5
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    Re: Solving Logarithmic Fractions

    Thanks that solved it. I wasn't aware of the identity you mentioned but it makes this problem quite simple and I think its the way we are expected to solve it. Thanks!
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  6. #6
    MHF Contributor MarkFL's Avatar
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    Re: Solving Logarithmic Fractions

    Continuing the method you began:

    \ln x\left(\frac{3}{2\ln 2}-\frac{5}{3\ln 2}\right)=-\frac{1}{2}

    \ln x\left(-\frac{1}{6\ln 2}\right)=-\frac{1}{2}

    \ln x=3\ln 2

    \ln x=\ln 2^3

    x=2^3

    Either method is probably what you are expected to know and use.
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