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Math Help - fractional indices

  1. #1
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    fractional indices

    can anyone help me with this question
    simplify

    9^-1/2 x 8^2/3 ?
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  2. #2
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    Quote Originally Posted by daniel123 View Post
    9^-1/2 x 8^2/3 ?
    9^{\frac{-1}{2}}\times 8^{\frac{2}{3}}

    \frac{1}{9^{\frac{1}{2}}}\times 8^{\frac{2}{3}}

    \frac{1}{\sqrt{9}}\times (\sqrt[3]{8})^2

    Can you finish it?
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  3. #3
    Senior Member eumyang's Avatar
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    9^{-1/2} \cdot 8^{2/3}

    First, if you have a negative exponent, you can make it positive by putting it in the denominator of a fraction:
    = \frac{1}{9^{1/2}} \cdot 8^{2/3}

    An exponent of 1/2 is the same as square root, so the square root of 9 is 3:
    = \frac{1}{3} \cdot 8^{2/3}

    The 8 to the 2/3 power can be rewritten using the power-of-a-power property:
    = \frac{1}{3} \cdot (8^{1/3})^2

    An exponent of 1/3 is the same as cube root, so the cube root of 8 is 2:
    = \frac{1}{3} \cdot 2^2

    Simplify:
    = \frac{4}{3}

    EDIT: Sorry pickslides!
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  4. #4
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    ok thankyou very much. I was going to leave it as square root of 9^-1 x cube root of 8^2
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