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Math Help - Evaluate 9^(-3/2)

  1. #1
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    Evaluate 9^(-3/2)

    Evaluate:

    9^(-3/2)

    Got a mental blank, please guys help me quickly!
    How is it 1/27? Please show complete working out.
    Last edited by mr fantastic; June 10th 2010 at 10:38 PM. Reason: Re-titled.
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  2. #2
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     <br />
9^{1/2}\;=\;3<br />
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  3. #3
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    9^(-3/2)

    9^(-3/2) = 1/(square root of 9^3 )
    = 1/27
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  5. #5
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    For any number a\ne 0, the number a^{-1} is defined as such a number that a\cdot{a^{-1}} =  a^{-1}\cdot{a} = 1. If we for example take 3, then the number 3^{-1} such that 3\cdot{3^{-1}} = 3^{-1}\cdot{3} = 1 is clearly \dfrac{1}{3}, as 3\cdot\left(\dfrac{1}{3}\right) = \left(\dfrac{1}{3}\right)\cdot{3} = 1. It can be easily seen that a^{-1} is always \dfrac{1}{a} (with the exception of a = 0, of course). So, for your case, we have 9^{-\frac{3}{2}} = \dfrac{1}{9^\frac{3}{2}} Remember that X^{\frac{Y}{Z}} = (X^{\frac{1}{Z}})^Y, so you have \dfrac{1}{9^\frac{3}{2}} = \dfrac{1}{(9^{\frac{1}{2}})^3}. Since 9^{\frac{1}{2}} = 3, we get \dfrac{1}{3^3}, which is of course 27.
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  6. #6
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    Quote Originally Posted by Joker37 View Post
    Evaluate:

    9^(-3/2)

    Got a mental blank, please guys help me quickly!
    How is it 1/27? Please show complete working out.
    You need to thoroughly review your basic index laws (and that's you job, not ours).

    A start (and please note the first reply you got):

    9^{-3/2} = \frac{1}{9^{3/2}} = \frac{1}{(9^{1/2})^3}.
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