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  1. #1
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    logs problem giving trouble

    3^{2-x}=2^{2x-1}
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  2. #2
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    Quote Originally Posted by math321 View Post
    3^{2-x}=2^{2x-1}
    \log(3^{2-x}) = \log(2^{2x-1})

    (2-x)\log{3} = (2x-1)\log{2}

    distribute both sides, get all terms with x on one side, then solve for x.
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  3. #3
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    Hi Math321

    This equation, when log laws are applied

    3^{2-x}=2^{2x-1}

    gives you,

    (2-x)\ln(3)=(2x-1)\ln(2)

    now expand each side, group like terms and finish.
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  4. #4
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    Quote Originally Posted by math321 View Post
    3^{2-x}=2^{2x-1}
    3^{2-x}=2^{2x-1}

    (2-x)\ln{3}= (2x-1)\ln{2}

    2\ln{3}-x\ln{3}= 2x\ln{2}-\ln{2}

    2\ln{3}+\ln{2}= 2x\ln{2}+x\ln{3}

    2\ln{3}+\ln{2}= x(2\ln{2}+\ln{3})

    \frac{2\ln{3}+\ln{2}}{2\ln{2}+\ln{3}}= x

     x = \frac{2\ln{3}+\ln{2}}{2\ln{2}+\ln{3}} = \frac{\ln18}{\ln{12}}
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  5. #5
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    Hello, math321!

    Solve for x\!:\;\; 3^{2-x}\:=\:2^{2x-1}

    Take logs: . \ln\left(3^{2-x}\right) \:=\:\ln\left(2^{2x-1}\right) \quad\Rightarrow\quad (2-x)\ln3 \:=\:(2x-1)\ln 2

    . . 2\ln 3 - x\ln 3 \:=\:2x\ln2 - \ln 2 \quad\Rightarrow\quad 2x\ln 2 + x\ln 3 \:=\:2\ln3 + \ln 2

    . . x(2\ln2 + \ln3) \:=\:2\ln3 + \ln 2 \quad\Rightarrow\quad x \:=\:\frac{2\ln 3 + \ln 2}{2\ln 2 + \ln 3}

    . . x \:=\:\frac{\ln(3^2) + \ln 2}{\ln(2^2) + \ln3} \:=\:\frac{\ln(9\cdot2)}{\ln(4\cdot3)} \:=\:\frac{\ln18}{\ln12}


    Therefore: . x \;\approx\;1.163171163

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