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Math Help - [SOLVED] Log proof

  1. #1
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    [SOLVED] Log proof

    If:

    6^{2x} = 3^{x+1}

    Prove that x = \;\log_{12}3

    Any tips on identity's and proving would be helpful thanks i seem to struggle with them.
    Last edited by MarcoMP; February 11th 2009 at 08:27 AM. Reason: Incorrect question
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  2. #2
    MHF Contributor red_dog's Avatar
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    This is not true.
    For x=1 we have 36=9!!!

    Maybe you have to solve the equation.
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  3. #3
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    My apologies red dog I left out the rest of the question but I have fixed it up now
    Last edited by MarcoMP; February 11th 2009 at 08:28 AM. Reason: mistake
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  4. #4
    MHF Contributor red_dog's Avatar
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    6^{2x}=3^{x+1}\Rightarrow 36^x=3\cdot 3^x\Rightarrow\log_{12}36^x=\log_{12}(3\cdot 3^x)\Rightarrow

    \Rightarrow x\log_{12}36=\log_{12}3+x\log_{12}3\Rightarrow x\left(\log_{12}36-\log_{12}3\right)=\log_{12}3\Rightarrow

    \Rightarrow x\cdot\log_{12}\frac{36}{3}=\log_{12}3\Rightarrow x\log_{12}12=\log_{12}3\Rightarrow x=\log_{12}3
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  5. #5
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    Another way:
    \begin{array}{rcl}<br />
   {6^{2x} } &  =  & {2^{2x}  \cdot 3^{2x} }  \\<br />
   {2^{2x}  \cdot 3^{2x} } &  =  & {3^{x + 1} }  \\<br />
   {2^{2x}  \cdot 3^x } &  =  & 3  \\<br />
   {12^x } &  =  & 3  \\<br />
   x &  =  & {\log _{12} (3)}  \\<br /> <br />
 \end{array}
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