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Math Help - Simplifying root denominators

  1. #1
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    Simplifying root denominators

    This expression I understand perfectly

    \frac{x}{\sqrt[4]x}=\frac{x}{\sqrt[4]x}*\frac{\sqrt[4]x^3}{\sqrt[4]x^3}

    = {\sqrt[4]x^3}

    This however I am completely lost...

    \frac{9}{\sqrt[5]8}=\frac{9}{\sqrt[5]2^3}=\frac{9}{\sqrt[5]2^3}*\frac{\sqrt[5]2^2}{\sqrt[5]2^2}=\frac{9\sqrt[5]4}{\sqrt[5]2^5}=\frac{9\sqrt[5]4}{2}

    Why must the  \sqrt[5]8 be broken down into \sqrt[5]2^3, and how are you supposed to know, and how is this done?

    I would solve it like this, why wouldnt this work?
    \frac{9}{\sqrt[5]8}=\frac{9}{\sqrt[5]8}*\frac{\sqrt[5]8^4}{\sqrt[5]8^4}=\frac{9\sqrt[5]4096}{\sqrt[5]8^4}=\frac{9\sqrt[5]4}{2}<br />
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  2. #2
    MHF Contributor Amer's Avatar
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    Quote Originally Posted by allyourbass2212 View Post
    This expression I understand perfectly

    \frac{x}{\sqrt[4]x}=\frac{x}{\sqrt[4]x}*\frac{\sqrt[4]x^3}{\sqrt[4]x^3}

    = {\sqrt[4]x^3}

    This however I am completely lost...

    \frac{9}{\sqrt[5]8}=\frac{9}{\sqrt[5]2^3}=\frac{9}{\sqrt[5]2^3}*\frac{\sqrt[5]2^2}{\sqrt[5]2^2}=\frac{9\sqrt[5]4}{\sqrt[5]2^5}=\frac{9\sqrt[5]4}{2}

    Why must the  \sqrt[5]8 be broken down into \sqrt[5]2^3, and how are you supposed to know, and how is this done?

    I would solve it like this, why wouldnt this work?
    \frac{9}{\sqrt[5]8}=\frac{9}{\sqrt[5]8}*\frac{\sqrt[5]8^4}{\sqrt[5]8^4}=\frac{9\sqrt[5]4096}{\sqrt[5]8^4}=\frac{9\sqrt[5]4}{2}<br />
    \frac{9}{\sqrt[5]{8}} = \frac{9}{\sqrt[5]{8}} \times \frac{\sqrt[5]{8^4}}{\sqrt[5]{8^4}}

    \frac{9\sqrt[5]{8^4}}{\sqrt[5]{8^5}} =\frac{9(\sqrt[5]{8^4})}{8}=\frac{9(\sqrt[5]{2^{12}})}{8}

    since \sqrt[5]{8^4}=\sqrt[5]{8*8*8*8}=\sqrt[5]{2^3(2^3)(2^3)(2^3)}=\sqrt[5]{(2^3)^4}= \sqrt[5]{2^{3(4)}}=\sqrt[5]{2^{12}}

    \frac{9(\sqrt[5]{2^{12}})}{8}=\frac{9\sqrt[5]{2^{10}(2^2)}}{8}=\frac{9(4)\sqrt[5]{4}}{8}=\frac{9\sqrt[5]{4}}{2}

    he broke

    \sqrt[5]{8} = \sqrt[5]{2^3} to make it easier so you will multiply with \frac{\sqrt[5]{2^2}}{\sqrt[5]{2^2}}

    but you see if we do not broke it it will be more harder ...
    Last edited by Amer; June 12th 2009 at 09:57 AM.
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