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Thread: Proving Horrible exponent

  1. #1
    Member BobBali's Avatar
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    Proving Horrible exponent

    Hi All,

    Prove

    $\displaystyle \frac{a}{ (\sqrt[3]{ax}}^2 = \frac{\sqrt[3]{ax}{x}$

    $\displaystyle \frac{a}{ (ax)^{1/3} )^2 = \frac{a}{(ax)^{2/3} }$

    $\displaystyle \frac{a}{ (ax)^{1/3} \times (ax)^{1/3}}$


    Had Latex trouble (again); question is in attachment below. Thanks
    Attached Thumbnails Attached Thumbnails Proving Horrible exponent-math.gif  
    Last edited by BobBali; Nov 23rd 2011 at 12:11 AM.
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  2. #2
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    Re: Proving Horrible exponent

    Quote Originally Posted by BobBali View Post
    Prove:

    $\displaystyle \frac{a}{\sqrt[3]{ax}^2} = \frac{\sqrt[3]{ax}}{x}$

    $\displaystyle \frac{a}{ \left((ax)^{1/3} \right)^2} = \frac{a}{ (ax)^{2/3} }}$

    $\displaystyle \frac{a}{ (ax)^{1/3} \times (ax)^{1/3}}$
    I fixed the LaTeX only! (Not sure if this was meant ....?)
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  3. #3
    Member BobBali's Avatar
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    Re: Proving Horrible exponent

    Yes it is! How did you do it? My latex working in the second step (the one u fixed) is below, can you tell me where i went wrong?? Thanks.

    $\displaystyle \frac{a}{ ( (\sqrt[3]{ax}) )^2 = \frac{a}{ ( (ax)^{1/3} )^2$
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    Re: Proving Horrible exponent

    Quote Originally Posted by BobBali View Post
    Yes it is! How did you do it? My latex working in the second step (the one u fixed) is below, can you tell me where i went wrong?? Thanks.

    $\displaystyle \frac{a}{ ( (\sqrt[3]{ax}) )^2 = \frac{a}{ ( (ax)^{1/3} )^2$
    \frac{a}{ \left(\sqrt[3]{ax} \right)^2 } = \frac{a}{ \left( (ax)^{1/3} \right)^2 }

    yields

    $\displaystyle \frac{a}{ \left(\sqrt[3]{ax} \right)^2 } = \frac{a}{ \left( (ax)^{1/3} \right)^2 }$

    I've marked the changes and additions in red.
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  5. #5
    Member BobBali's Avatar
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    Re: Proving Horrible exponent

    Ok, so:

    $\displaystyle \frac{a}{ \left (ax)^{2/3} \right }$ =
    $\displaystyle \frac{a}{ \left(ax)^{1/3} \times (ax)^{1/3} \right)$

    $\displaystyle a \times a^{-1/3} \times x^{-1/3} \times a^{-1/3} \times x^{-1/3} =$
    $\displaystyle a^{1/3} \times x^{-2/3}$ =
    $\displaystyle \frac{\sqrt[3]{a}{x}^{2/3} $

    Still doesn't yield answer above?
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  6. #6
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    Re: Proving Horrible exponent

    Quote Originally Posted by BobBali View Post
    Ok, so:

    $\displaystyle \frac{a}{ \left (ax)^{2/3} \right }$ =
    $\displaystyle \frac{a}{ \left(ax)^{1/3} \times (ax)^{1/3} \right)}$

    $\displaystyle a \times a^{-1/3} \times x^{-1/3} \times a^{-1/3} \times x^{-1/3} =$
    $\displaystyle a^{1/3} \times x^{-2/3}$ =
    $\displaystyle \frac{\sqrt[3]{a}}{x^{2/3}} $

    Still doesn't yield answer above?
    Please use the \frac-command in such a way:

    \frac{numerator}{denominator} yields $\displaystyle \frac{numerator}{denominator}$
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  7. #7
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    Re: Proving Horrible exponent

    Hello, BobBali!

    $\displaystyle \text{Prove: }\:\frac{a}{(\sqrt[3]{ax})^2} \:=\:\frac{\sqrt[3]{ax}}{x}$

    On the left side, we have: .$\displaystyle \frac{a}{(ax)^{\frac{2}{3}}}$

    Multiply by $\displaystyle \tfrac{(ax)^{\frac{1}{3}}}{(ax)^{\frac{1}{3}}}: \;\;\frac{a}{(ax)^{\frac{2}{3}}} \cdot\frac{(ax)^{\frac{1}{3}}}{(ax)^{\frac{1}{3}}} \;=\;\frac{a(ax)^{\frac{1}{3}}}{ax} \;=\;\frac{(ax)^{\frac{1}{3}}}{x} \;=\;\frac{\sqrt[3]{ax}}{x} $

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