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Math Help - Solving the unknown base of exponents

  1. #1
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    Solving the unknown base of exponents

    So i realised that exponents are the same thing as indices which I put down to the different countries that we all come from.

    Anyway - I have this question which i didn't really know how to approach:

    solve for 'x':

    x^3/2 = 8^1/2 * 9^1/3 / 6^4/3


    I did some working and got:
    x^3/2 = 2^3/4 * 3^-1/4

    I'm not sure if my working was right, but even so, I don't know how to proceed with this question.
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  2. #2
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    Hello, Ogsta101!

    Solve for x\!:\;\;x^{\frac{3}{2}} \;=\;\frac{8^{\frac{1}{2}}\cdot9^{\frac{1}{3}}}{6^  {\frac{4}{3}}}
    Note that all the bases have factors of 2 or 3.

    We have: . x^{\frac{3}{2}} \;=\;\frac{\left(2^3\right)^{\frac{1}{2}}\cdot\lef  t(3^2\right)^{\frac{1}{3}}} {(2\cdot3)^{\frac{4}{3}}} \;=\;\frac{2^{\frac{3}{2}}\cdot3^{\frac{2}{3}}}{2^  {\frac{4}{3}}\cdot3^{\frac{4}{3}}}\quad\Rightarrow  \quad x^{\frac{3}{2}}  \;=\;\frac{2^{\frac{1}{6}}}{3^{\frac{2}{3}}}

    Raise both sides to the power \frac{2}{3}\!:\;\;\left(x^{\frac{3}{2}}\right)^{\f  rac{2}{3}} \;=\;\left(\frac{2^{\frac{1}{6}}}{3^{\frac{2}{3}}}  \right)^{\frac{2}{3}}

    Therefore: . \boxed{x\;=\;\frac{2^{\frac{1}{9}}}{3^{\frac{4}{9}  }} }

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