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Thread: 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.
    $\displaystyle \log_a{16} = \log_a{(2^4)}$

    $\displaystyle = 4\log_a{2}$.

    Therefore $\displaystyle x = 4\log_a{2}$

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


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

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

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

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



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

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

    We can know that:
    $\displaystyle
    a=\sqrt[y]{12}$ or
    $\displaystyle a=\sqrt[x]{16}$

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

    $\displaystyle a=\sqrt[\delta]{3}$


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

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

    $\displaystyle
    \delta=\textrm{log}_{12^{\frac{1}{y}}}3\,\,\,\,\,\ therefore$
    $\displaystyle
    \delta=y\textrm{log}_{12}3\,\,\,\,\because log_{x^n}y=\frac{log_xy}{n}$

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

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

    Therefore:

    $\displaystyle y=log_a12$


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

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

    $\displaystyle \delta=xlog_{16}3$

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

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

    $\displaystyle x=4log_a2$


    Test:

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

    &
    x=12log_32
    So

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

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