Indices simplification

**gary223** · August 2nd 2007, 11:59 PM

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}

**tukeywilliams** · August 3rd 2007, 12:28 AM

$\sqrt{\frac{81a^6 \cdot b^2 \cdot c^{-4}}{36a^2 \cdot b^6 \cdot c^4}}$

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

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

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

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

Rewriting in positive exponents we get: $\frac{9a^2}{4b^{2}c^{4}}$ .

**gary223** · August 3rd 2007, 02:16 AM

How would you do this one?

\sqrt[3]27X^8.Y.Z^-2/8X^2.Y^-2.Z^10

**topsquark** · August 3rd 2007, 06:10 AM

Originally Posted by gary223

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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