# simplification of exponential expression

• April 1st 2010, 05:48 PM
dymand68
simplification of exponential expression
[(x^2 y^-2)^-1]^-1=
(1/x^2 y^-2)^-1=
(y^2/x^2)-1=
x^2/y^2
this solution is right according to the back of the book, are the steps also correct?
• April 1st 2010, 05:55 PM
Prove It
Quote:

Originally Posted by dymand68
[(x^2y^-2)^-1]^-1=
(1/x^2y^-2)^-1=
(y^2/x^2)-1=
x^2/y^2
this solution is right according to the back of the book, are the steps also correct?

$[(x^2y^{-2})^{-1}]^{-1}$

$= (x^2y^{-2})^1$

$= x^2y^{-2}$

$= \frac{x^2}{y^2}$

$= \left(\frac{x}{y}\right)^2$.
• April 1st 2010, 05:59 PM
dymand68
ty.........so the correct order of operations starts with the negative exponent OUTSIDE the brackets?