I am having a problem with manipulating negaticve exponents the problem is 1/2x^-2y^-3 I come up with 2x^2y^3 but that does not seem right. Any thoughts Thanks JS
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Originally Posted by jsto I am having a problem with manipulating negaticve exponents the problem is 1/2x^-2y^-3 I come up with 2x^2y^3 but that does not seem right. Any thoughts Thanks JS Just making sure...is it $\displaystyle \frac{1}{2x^{-2}y^{-3}}$??
Originally Posted by Chris L T521 Just making sure...is it $\displaystyle \frac{1}{2x^{-2}y^{-3}}$?? Yes that's right
EDIT: Cocked it up, see correct answer below
Last edited by rowe; Oct 12th 2009 at 11:07 AM.
Originally Posted by rowe $\displaystyle \frac{1}{2x^{-2}y^{-3}} = \frac{1}{\frac{1}{2x^2y^3}}=1 \cdot \frac{2x^2y^3}{1} = 2x^2y^3$ I don't want to pick at you but the 2 in the denominator has NOT a negative exponent $\displaystyle \frac{1}{2x^{-2}y^{-3}} = \frac{1}{2 \cdot \frac{1}{x^2y^3}}= \dfrac{x^2y^3}2$
Originally Posted by rowe $\displaystyle \frac{1}{2x^{-2}y^{-3}} = \frac{1}{\frac{1}{2x^2y^3}}=1 \cdot \frac{2x^2y^3}{1} = 2x^2y^3$ The 2 is not exponentiated so you're a factor of 4 too high. $\displaystyle \frac{1}{2}x^2y^3$
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