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Math Help - Logarithms

  1. #1
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    Logarithms

    Given that log a 16= x and log a 12 = y, express loga3 in terms of x and y.
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  2. #2
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    Quote Originally Posted by Sxtxs View Post
    Given that log a 16= x and log a 12 = y, express loga3 in terms of x and y.
    \log_a{16} = \log_a{(2^4)}

     = 4\log_a{2}.

    Therefore x = 4\log_a{2}

    2\log_a{2} = \frac{x}{2}.


    \log_a{12} = \log_a{(3\cdot 4)}

     = \log_a{3} + \log_a{2^2}.

    Therefore y = \log_a{3} + 2\log_a{2}

    y = \log_a{3} + \frac{x}{2}.



    So \log_a{3} = y - \frac{x}{2}.
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  3. #3
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    Suppose:
    \textrm{log}_a16=x
    \textrm{log}_a12=y

    Express  \textrm{log}_a3 in terms of x&y.

    We can know that:
    <br />
a=\sqrt[y]{12} or
    a=\sqrt[x]{16}

    We can also infer, if log_a3=\delta that:

    a=\sqrt[\delta]{3}


    So we have:
    \sqrt[\delta]{3}=\sqrt[x]{16}=\sqrt[y]{12} and

    \sqrt[\delta]{3}=\sqrt[y]{12}\,\,\,\,\,\,\therefore

    <br />
\delta=\textrm{log}_{12^{\frac{1}{y}}}3\,\,\,\,\,\  therefore
    <br />
\delta=y\textrm{log}_{12}3\,\,\,\,\because log_{x^n}y=\frac{log_xy}{n}

    y=\frac{\delta}{log_{12}3}

    y=\frac{log_a3}{log_{12}3}

    Therefore:

    y=log_a12


    Like wise:
    \sqrt[\delta]{3}=\sqrt[x]{16}

    \delta=log_{16^{\frac{1}{x}}}{3}

    \delta=xlog_{16}3

    x=\frac{\delta}{log_{16}3}

    x=\frac{log_a3}{log_{16}3}

    x=4log_a2


    Test:

    If
    \delta=3
    then
    y=3log_312
    so
    \frac{3}{log_{12}3}=3log_312

    &
    x=12log_32
    So

    x=\frac{3}{log_{16}3}=12log_32

    The test values where derived from:
    \sqrt[\delta]{3}=\sqrt[x]{16}=\sqrt[y]{12}
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