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Math Help - Indices simplification

  1. #1
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    Indices simplification

    Use the laws of indices to simplify each of the following, giving answers with positive indices only.

    SQRL{81A^6.B^2.C^-4/36A^2.B^6.C^4}
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  2. #2
    Senior Member tukeywilliams's Avatar
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     \sqrt{\frac{81a^6 \cdot b^2 \cdot c^{-4}}{36a^2 \cdot b^6 \cdot c^4}}

     \sqrt{\frac{9a^6 \cdot b^2 \cdot c^{-4}}{4a^2 \cdot b^6 \cdot c^4}}

     \left(\frac{9a^6 \cdot b^2 \cdot c^{-4}}{4a^2 \cdot b^6 \cdot c^4} \right)^{1/2}

     \frac{9a^3 \cdot b \cdot c^{-2}}{4a \cdot b^3 \cdot c^2}

    or  \frac{9}{4}a^{2} \cdot b^{-2} \cdot c^{-4}

    Rewriting in positive exponents we get:  \frac{9a^2}{4b^{2}c^{4}} .
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  3. #3
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    How would you do this one?

    \sqrt[3]27X^8.Y.Z^-2/8X^2.Y^-2.Z^10
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by gary223 View Post
    How would you do this one?

    \sqrt[3]27X^8.Y.Z^-2/8X^2.Y^-2.Z^10
    New questions should go in new threads.

    \sqrt[3]{\frac{27x^8yz^{-2}}{8x^2y^{-2}z^{10}}}

    =\sqrt[3]{\frac{27x^8y\frac{1}{z^2}}{8x^2\frac{1}{y^2}z^{10  }}}

    =\sqrt[3]{\frac{27x^8yy^2}{8x^2z^{10}z^2}}

    =\sqrt[3]{\frac{27x^8y^3}{8x^2z^{12}}}

    =\sqrt[3]{\frac{3^3(x^3)^2x^2y^3}{2^3x^2(z^3)^4}}

    =\frac{3x^2y}{2z^4}\sqrt[3]{\frac{x^2}{x^2}}

    =\frac{3x^2y}{2z^4}

    -Dan
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