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Math Help - Solving equations with indeces (version 2)

  1. #1
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    Solving equations with indeces (version 2)

    Hello everyone,

    I have come undone on solving another equation with indices in:

    \frac{3^{5x+2}}{9^{1-x}}=\frac{27^{4+3x}}{729}

    I've a feeling it involves simplifying the numerator and denominator of the second half so that I can multiply with the first half, but I'm doing something wrong.

    Help much appreciated.
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    Re: Solving equations with indeces (version 2)

    Quote Originally Posted by mabentley View Post
    \frac{3^{5x+2}}{9^{1-x}}=\frac{27^{4+3x}}{729}
    9^{1-x}=3^{2-2x},~~27^{4+3x}=3^{12+9x},~~727=3^6
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  3. #3
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    Re: Solving equations with indeces (version 2)

    These are all powers of three: 27 = 3^3 ,\ 9 = 3^2\ ,\ 729 = 3^6

    Hence the question is: \dfrac{3^{5x+2}}{3^{2(1-x)}} = \dfrac{3^{3(4+3x)}}{3^6}

    Use the laws of indices to simplify
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