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Math Help - Logarithmic equations

  1. #1
    Member Furyan's Avatar
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    Logarithmic equations

    Hello

    The question is, solve:

    2\log_{x}4 - \log_{2}x = 3

    Changing the base I get:

    \log_{x}16 - \dfrac{\log_{x}x}{\log_{x}2} = 3

    \log_{x}16 - \dfrac{1}{\log_{x}2} = 3

    I don't know how to proceed from here.

    I know that the difference between two logs with the same base can be written as a division, but the reciprocal is throwing me off.

    I would greatly appreciate some help.

    Thank you.
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  2. #2
    MHF Contributor
    skeeter's Avatar
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    Re: Logarithmic equations

    2\log_x{4} - \log_2{x} = 3

    2 \cdot \frac{\log_2{4}}{\log_2{x}} - \log_2{x} = 3

    \frac{4}{\log_2{x}} - \log_2{x} = 3

    let u = \log_2{x} ...

    \frac{4}{u} - u = 3

    4 - u^2 = 3u

    0 = u^2 + 3u - 4

    solve the quadratic for u , then back substitute to solve for x.
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  3. #3
    Member Furyan's Avatar
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    Re: Logarithmic equations

    Hello skeeter,

    I had also tried changing the base of the log to 2 and got:

    \frac{4}{\log_2{x}} - \log_2{x} = 3

    but I hadn't thought of using a substitution and then solving.

    Brilliant! Thank you very much.
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