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Math Help - Negative radical exponent simplification

  1. #1
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    Negative radical exponent simplification

    Hi,

    I am dealing with the following problem and am unsure whether
    I solved it right:
    =================================================
    (16xy)^(1/4) (-8x^2z^3)^(-1/3)
    --------------------------------
    (25^-1 x^2 yz)^(-1/2)

    2(xy)^(1/4) (-8x^2z^3)^(-1/3)
    = ------------------------------
    ((25^-1 x^2 yz)^-1)^(1/2))

    2(xy)^(1/4) (-8x^2z^3)^(-1/3)
    = --------------------------------
    (25x^-2y^-1z^-1)^(1/2)

    2(xy)^(1/4)
    = ---------------------------------------------
    5(x^-2y^-1z^-1)^(1/2) (-8x^2z^3)^(1/3)


    = 2(xy)^(1/4)
    ---------------------------------
    5(x^-2y^-1z^-1)^(1/2) -2z(x^2)^(1/3)
    ===================================

    Thanks for the help,

    Alex
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  2. #2
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    \displaystyle\frac{(16xy)^{1/4}*(-8x^2z^3)^{-1/3}}{(25^{-1}x^2yz)^{-1/2}}

    Is this is the expression?
    Last edited by dwsmith; January 6th 2011 at 12:55 PM. Reason: Inserted a (-)
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  3. #3
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    Yes sir
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  4. #4
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    Actually, the 1/2 exponent in the denominator is negative 1/2 (-1/2)
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  5. #5
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    It did it on the fly, by I get

    \displaystyle -x^{\frac{7}{12}}y^{\frac{3}{4}}z^{\frac{-2}{3}}
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  6. #6
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    Would you mind showing the steps for my learning purposes?

    Thanks,

    Alex
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  7. #7
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    Quote Originally Posted by alex95 View Post
    Would you mind showing the steps for my learning purposes?
    No problem. Where possible for all terms

    Step 1: apply \displaystyle (a^m)^n = a^{m\times n}

    Step 2: apply \displaystyle a^m\times a^n = a^{m+ n}

    Step 3: apply \displaystyle a^m \div a^n = a^{m- n}
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  8. #8
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    Hello, alex95!

    \displaystyle\text{SimpliftyL } \;\frac{(16xy)^{\frac{1}{4}}(-8x^2z^3)^{-\frac{1}{3}}} {(2s^{-1}x^2yz)^{-\frac{1}{2}}}

    We have: . \displaystyle \frac{(16xy)^{\frac{1}{4}}(2s^{-1}x^2yz)^{\frac{1}{2}}} {(-8x^2z^3)^{\frac{1}{3}}}

    . . . . . . \displaystyle =\; \frac{16^{\frac{1}{4}}\cdot x^{\frac{1}{4}}\cdot y^{\frac{1}{4}} \cdot 2^{\frac{1}{2}}\cdot(s^{-1})^{\frac{1}{2}}\cdot(x^2)^{\frac{1}{2}}\cdot y^{\frac{1}{2}}\cdot z^{\frac{1}{2}}} {(-8)^{\frac{1}{3}}(x^2)^{\frac{1}{3}}(z^3)^{\frac{1}  {3}}}

    . . . . . . \displaystyle =\;\frac{2\cdot x^{\frac{1}{4}}\cdot y^{\frac{1}{4}}\cdot 2^{\frac{1}{2}}\cdot x\cdot y^{\frac{1}{2}}\cdot z^{\frac{1}{2}}\cdot s^{-\frac{1}{2}}}  {-2\cdot x^{\frac{2}{3}}\cdot z}

    . . . . . . \displaystyle =\; \frac{2^{\frac{3}{2}}\cdot x^{\frac{5}{4}} \cdot y^{\frac{3}{4}}\cdot z^{\frac{1}{2}}\cdot s^{-\frac{1}{2}}} {-2\cdot x^{\frac{2}{3}}\cdot z}

    . . . . . . \displaystyle\;=-\frac{2^{\frac{1}{2}}\cdot x^{\frac{7}{12}}\cdot y^{\frac{3}{4}}} {z^{\frac{1}{2}}\cdot s^{\frac{1}{2}}}

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  9. #9
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    \displaystyle\frac{(16xy)^{\frac{1}{4}}\;\left(-8x^2z^3\right)^{-\frac{1}{3}}}{\left(25^{-1}x^2yz\right)^{-\frac{1}{2}}}=\frac{16^{\frac{1}{4}}x^{\frac{1}{4}  }y^{\frac{1}{4}}\;\left(25^{-1}x^2yz\right)^{\frac{1}{2}}}{\left(-8x^2z^3\right)^{\frac{1}{3}}}


    =\displaystyle\frac{2x^{\frac{1}{4}}y^{\frac{1}{4}  }xy^{\frac{1}{2}}z^{\frac{1}{2}}}{-2zx^{\frac{2}{3}}25^{\frac{1}{2}}}=\frac{x^{\frac{  1}{4}}x^1y^{\frac{1}{4}}y^{\frac{1}{2}}z^{\frac{1}  {2}}}{-5x^{\frac{2}{3}}z}


    =\displaystyle\ -\frac{x^{\left(\frac{5}{4}-\frac{2}{3}\right)}y^{\frac{3}{4}}z^{\left(\frac{1  }{2}-1\right)}}{5}


    =-\displaystyle\frac{x^{\frac{7}{12}}y^{\frac{3}{4}}  }{5z^{\frac{1}{2}}}
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  10. #10
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    Quote Originally Posted by Soroban View Post
    \displaystyle\;=-\frac{2^{\frac{1}{2}}\cdot x^{\frac{7}{12}}\cdot y^{\frac{3}{4}}} {z^{\frac{1}{2}}\cdot s^{\frac{1}{2}}}
    Another great post Soroban, but I don't see where you found an 's' in all of that... I think Archie has it nailed.
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