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Thread: Unable to solve roots expression

  1. #16
    MHF Contributor Amer's Avatar
    May 2009
    ok all I can do is to give examples with little explanation

    any number you can write it in power of 1

    $\displaystyle x=x^{1}$

    $\displaystyle y^2=(y^2)^{1} $

    $\displaystyle \frac{1}{x^2}=\frac{1}{(x^2)^{1}}$ so if I would to move the denominator to the numerator you will have

    $\displaystyle \frac{1}{x^2}=\frac{1}{(x^2)^{1}}=(x^2)^{-1} = x^{2(-1)}=x^{-2}$

    I think it is ok for now right....

    $\displaystyle \sqrt[3]{x}=\sqrt[3]{x^{1}} = (x^{1})^{\frac{1}{3}}=x^{1(\frac{1}{3})}=x^{\frac{ 1}{3}}$

    $\displaystyle \frac{1}{\sqrt[3]{x}} = \frac{1}{\sqrt[3]{x^{1}}}=\frac{1}{(x^{1})^{\frac{1}{3}}}=\frac{1}{ x^{1(\frac{1}{3})}}=\frac{1}{x^{\frac{1}{3}}}=x^{(-1)(\frac{1}{3})}=x^{\frac{-1}{3}}$

    $\displaystyle \sqrt{xy} = (xy)^\frac{1}{2} $ this is clear right

    $\displaystyle \frac{1}{\sqrt{xy}} = \frac{1}{(xy)^{\frac{1}{2}}} = ((xy)^{\frac{1}{2}})^{-1} = (xy)^{(-1)(\frac{1}{2})}=(xy)^{\frac{-1}{2}}$

    $\displaystyle \frac{1}{\sqrt[3]{xy}}=\frac{1}{(xy)^{\frac{1}{3}}}=\left((xy)^{\fr ac{1}{3}}\right)^{-1} = (xy)^{(\frac{1}{3})(-1)}=(xy)^{\frac{-1}{3}} $

    $\displaystyle \sqrt[5]{x^2y} = \sqrt[5]{(x^2y)^{1}} = \left((x^2y)^{1}\right)^{\frac{1}{5}} = (x^2y)^{(1)(\frac{1}{5})} = (x^2y)^{\frac{1}{5}}$

    $\displaystyle \frac{1}{\sqrt[5]{x^2y}}=\frac{1}{\sqrt[5]{(x^2y)^{1}}}=$$\displaystyle \frac{1}{((x^2y)^{1})^{\frac{1}{5}}}=\frac{1}{(x^2 y)^{\frac{1}{5}}}=((x^2y)^{\frac{1}{5}})^{-1}=$$\displaystyle (x^2y)^{\left(\frac{1}{5}\right)(-1)}=(x^2y)^{\frac{-1}{5}}$

    it is clear or not , is there anything else or anything is not clear ?
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  2. #17
    May 2008
    Thanks Amers the examples are helpful, but to make sure I got the hang of it I created several examples to see if I created them as exponents correctly.

    $\displaystyle \frac{1}{\sqrt[5]{3xy^7}}= (3xy^7)^{-1/5}$

    $\displaystyle \frac{1}{\sqrt[8]{2xyz^7}}=(2xyz^7)^{-1/8}$

    $\displaystyle \sqrt[5]{x^4}=x^{4/5}$

    $\displaystyle \sqrt[7]{7x^4}=(7x^4)^{1/7}$

    $\displaystyle \sqrt[5]{(7x-4)^3}=(7x-4)^{3/5}$
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