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Thread: 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; Jun 10th 2010 at 10:38 PM. Reason: Re-titled.
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  2. #2
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    $\displaystyle
    9^{1/2}\;=\;3
    $
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  3. #3
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    Negative Exponent Intuition



    [YOUTUBE]http://www.youtube.com/watch?v=Tqpcku0hrPU[/YOUTUBE]


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  4. #4
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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 $\displaystyle a\ne 0$, the number $\displaystyle a^{-1}$ is defined as such a number that $\displaystyle a\cdot{a^{-1}} = a^{-1}\cdot{a} = 1.$ If we for example take $\displaystyle 3$, then the number $\displaystyle 3^{-1}$ such that $\displaystyle 3\cdot{3^{-1}} = 3^{-1}\cdot{3} = 1 $ is clearly $\displaystyle \dfrac{1}{3}$, as $\displaystyle 3\cdot\left(\dfrac{1}{3}\right) = \left(\dfrac{1}{3}\right)\cdot{3} = 1.$ It can be easily seen that $\displaystyle a^{-1}$ is always $\displaystyle \dfrac{1}{a}$ (with the exception of $\displaystyle a = 0$, of course). So, for your case, we have $\displaystyle 9^{-\frac{3}{2}} = \dfrac{1}{9^\frac{3}{2}}$ Remember that $\displaystyle X^{\frac{Y}{Z}} = (X^{\frac{1}{Z}})^Y$, so you have $\displaystyle \dfrac{1}{9^\frac{3}{2}} = \dfrac{1}{(9^{\frac{1}{2}})^3}.$ Since $\displaystyle 9^{\frac{1}{2}} = 3$, we get $\displaystyle \dfrac{1}{3^3},$ which is of course $\displaystyle 27$.
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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):

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