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Thread: solve for x???

  1. September 12th 2010, 04:02 AM #1
    lollilikeslife
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    solve for x???

    3^(x+1)=7^(x-4)

    () are exponents


    I dont really know what to do? I think we convert each side into a log: (x+1)Log3 = (x-4)Log7

    But, I'm not sure? Please help! And provide work, I would really love to learn how to do this
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  2. September 12th 2010, 04:06 AM #2
    Prove It
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    That's exactly what you do.

    I usually use natural logarithms...


    3^{x + 1} = 7^{x - 4}

    \ln{\left(3^{x + 1}\right)} = \ln{\left(7^{x - 4}\right)}

    (x + 1)\ln{3} = (x - 4)\ln{7}

    x\ln{3} + \ln{3} = x\ln{7} - 4\ln{7}

    4\ln{7} + \ln{3} = x\ln{7} - x\ln{3}

    4\ln{7} + \ln{3} = x(\ln{7} - \ln{3})

    x = \frac{4\ln{7} + \ln{3}}{\ln{7} - \ln{3}}.
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