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Math Help - Logarithm - please help

  1. #1
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    Logarithm - please help

    Please shed some light on the following......

    If x and y are both positive and log_9 x = log_12 y = log_16 (x + y), find the value of x/y.

    Note: log_9 x should read as log base 9 x

    Thanks.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by madaboutmath View Post
    Please shed some light on the following......

    If x and y are both positive and log_9 x = log_12 y = log_16 (x + y), find the value of x/y.

    Note: log_9 x should read as log base 9 x

    Thanks.

    Put:

    u=\log_9(x)=\log_{12}(y)=\log_{16}(x+y)

    Then:

     9^u=x,\ 12^u=y,\ 16^u=x+y

    So:

    <br />
\frac{x+y}{y}=\left[\frac{16}{12}\right]^u=\left[\frac{4}{3}\right]^u<br />

    and:

    <br />
\frac{y}{x}=\left[\frac{12}{9}\right]^u=\left[\frac{4}{3}\right]^u<br />

    Therefore if z=x/y, we have:

    <br />
z+1=1/z<br />

    which is a quadratic in z, one (or more of which's solutions)
    should solve the problem.

    RonL

    RonL
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  3. #3
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    Logarithm

    I really appreciate your prompt help CaptainBlack. The answer is (sqrt5-1)/2.

    Thanks.

    MAM
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