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Math Help - Proving Horrible exponent

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

    Hi All,

    Prove

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

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

    \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; November 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:

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

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

    \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
    Junior 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.

    \frac{a}{ ( (\sqrt[3]{ax}) )^2     =       \frac{a}{ ( (ax)^{1/3} )^2
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  4. #4
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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.

    \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

    \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
    Junior Member BobBali's Avatar
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    Re: Proving Horrible exponent

    Ok, so:

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

    a \times a^{-1/3} \times x^{-1/3} \times a^{-1/3} \times x^{-1/3}  =
    a^{1/3} \times x^{-2/3} =
    \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:

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

    a \times a^{-1/3} \times x^{-1/3} \times a^{-1/3} \times x^{-1/3}  =
    a^{1/3} \times x^{-2/3} =
    \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 \frac{numerator}{denominator}
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  7. #7
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    Re: Proving Horrible exponent

    Hello, BobBali!

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

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

    Multiply by \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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