# negative exponents

• Oct 12th 2009, 09:25 AM
jsto
negative exponents
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
• Oct 12th 2009, 09:33 AM
Chris L T521
Quote:

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 $\frac{1}{2x^{-2}y^{-3}}$??
• Oct 12th 2009, 10:00 AM
jsto
Quote:

Originally Posted by Chris L T521
Just making sure...is it $\frac{1}{2x^{-2}y^{-3}}$??

Yes that's right
• Oct 12th 2009, 10:53 AM
rowe
EDIT: Cocked it up, see correct answer below
• Oct 12th 2009, 10:58 AM
earboth
Quote:

Originally Posted by rowe
$\frac{1}{2x^{-2}y^{-3}} = \frac{1}{\frac{1}{2x^2y^3}}=1 \cdot \frac{2x^2y^3}{1} = 2x^2y^3$

I don't want to pick at you but the 2 in the denominator has NOT a negative exponent

$\frac{1}{2x^{-2}y^{-3}} = \frac{1}{2 \cdot \frac{1}{x^2y^3}}= \dfrac{x^2y^3}2$
• Oct 12th 2009, 10:59 AM
e^(i*pi)
Quote:

Originally Posted by rowe
$\frac{1}{2x^{-2}y^{-3}} = \frac{1}{\frac{1}{2x^2y^3}}=1 \cdot \frac{2x^2y^3}{1} = 2x^2y^3$

The 2 is not exponentiated so you're a factor of 4 too high.

$\frac{1}{2}x^2y^3$