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Math Help - Logarithms... I think?

  1. #1
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    Logarithms... I think?

    Solve for x....

    2^(2x+1) = 3^(2x-5)
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  2. #2
    MHF Contributor
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    West Malaysia
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    Quote Originally Posted by CantcountWontcount View Post
    Solve for x....

    2^(2x+1) = 3^(2x-5)
    HI

    Apply log to both sides ,

     <br />
\log 2^{2x+1}=\log 3^{2x-5}<br />

    (2x+1)\log 2=(2x-5)\log 3

    \frac{2x+1}{2x-5}=\frac{\log 3}{\log 2}

    \frac{2x+1}{2x-5}=1.585

    2x+1=1.585(2x-5)

    Can you take it from here ?
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  3. #3
    Newbie
    Joined
    Oct 2009
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    Thanks addict, the answer on my tutorial sheet must be wrong...

    I completed it by doing

    (2x + 1)log2=(2x - 5)log3
    2xlog2+log2=2xlog3-5log3
    2xlog3-2xlog2=log2+5log3
    2x(log3-log2)=log2+5log3
    2x=(log2+5log3)/(log3-log2)......

    I get the same answer doing it this way, was pulling my hair out for ages.

    Thanks again
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