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Math Help - Simplify involving indices

  1. #1
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    Simplify involving indices

    Hi. I don't know the answer to this one, could someone check my reasoning and working please?

    Simplify \frac{2a^-2}{a^\frac{-3}{2}}

    Working on the denominator only
    \frac{1}{-\sqrt{a}*-\sqrt{a}*-\sqrt{a}}


    Denominator is \frac{1}{a\sqrt{a}}

    So I now have \frac{2a^-2}{1}*\frac{a\sqrt{a}}{1}

    Then \frac{1}{2a^2}*a\sqrt{a}

    Which gives me my answer \frac{a\sqrt{a}}{2a^2}
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  2. #2
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    \frac{2a^{-2}}{a^{\frac{-3}{2}}}= 2 \frac{a^{-2}}{a^{\frac{-3}{2}}}

    = 2 a^{{-2}-({\frac{-3}{2}})}= 2 a^{\frac{-4+3}{2}}

    = 2 a^{\frac{-1}{2}}  = \frac {2}{\sqrt a }

    or

    2 a^{-2} \times a^{\frac{3}{2}}

    2 a^{-2+\frac{3}{2}} = = 2 a^{\frac{-4+3}{2}}
    Last edited by ramiee2010; October 2nd 2009 at 08:04 AM. Reason: giving alternative method
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  3. #3
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    Working on the denominator only
    \frac{1}{-\sqrt{a}*-\sqrt{a}*-\sqrt{a}}
    a^{\frac{-3}{2}} = \frac{1}{a^{\frac{3}{2}}}=\frac{1}{\sqrt {a} \times \sqrt{ a} \times \sqrt {a}}
    \frac{2a^-2}{1}*\frac{a\sqrt{a}}{1}
    \frac{2}{a^2}*a\sqrt{a}

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  4. #4
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    Thank you. Was I correct in my thinking to here?

    even though it is probably not the usual way of dealing with this kind of question?

    and is \frac{2}{a^2}*a\sqrt{a} the final and correct answer to this question please?
    Last edited by Meggomumsie; October 2nd 2009 at 08:20 AM. Reason: Left something out
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  5. #5
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    \frac{2a^-2}{a^\frac{-3}{2}}
    2a^{-2}*a^\frac{3}{2}
    2a^{-2+\frac{3}{2}}
    2a^{-0.5}

    or

    \frac{2}{\sqrt{a}}
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