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Math Help - Challenging question on logarithm

  1. #1
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    Challenging question on logarithm

    Given that log_{9}x=log_{12}y=log_{16}(x+y), find \frac{x}{y}

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  2. #2
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    z = \frac{y}{x}

    log_{9}x = log_{12}xz = log_{16}x(1+z)

     <br />
\frac{lnx}{ln9} = \frac{lnx + lnz}{ln12} = \frac{lnx + ln(1+z)}{ln16}<br />

     <br />
lnx + lnx + ln(1+z) = 2(lnx + lnz)<br />
     <br />
ln(1+z) = 2 lnz<br />
    1+z = z^2

    z^2 - z - 1 = 0

    z = \frac{1+\sqrt{5}}{2} = \frac{y}{x}


    \frac{x}{y} = \frac{2}{1+\sqrt{5}}


    Patrick
    Last edited by PatrickFoster; October 15th 2009 at 09:37 PM. Reason: Typo
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  3. #3
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    Hello acc100jt
    Quote Originally Posted by acc100jt View Post
    Given that log_{9}x=log_{12}y=log_{16}(x+y), find \frac{x}{y}

    Thanks
    In case you don't follow how Patrick went from here:
      \frac{lnx}{ln9} = \frac{lnx + lnz}{ln12} = \frac{lnx + ln(1+z)}{ln16}
    to here:
      lnx + lnx + ln(1+z) = 2(lnx + lnz)
    ... here's an alternative approach, that just uses the fundamental property of a logarithm, namely
    a=\log_bc \iff b^a=c\quad (c>0)
    \log_{9}x=\log_{12}y=\log_{16}(x+y)=p, say

    \Rightarrow
    9^p=x
    12^p=y\Rightarrow144^p=y^2
    16^p=(x+y)
    Then, since 9^p \times 16^p=144^p:
    x(x+y)=y^2
    \Rightarrow \left(\frac{x}{y}\right)^2+\left(\frac{x}{y}\right  )-1=0

    \Rightarrow \frac{x}{y}=\frac{-1+\sqrt5}{2}\,\Big(=\frac{2}{1+\sqrt5}\Big), taking the positive root, since x>0,y>0

    Grandad
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