Hello SSK Originally Posted by
SSK (2x^-2 y^3/6xz^5)^-4
Answer is 81x^12 z^20/y^12
how is this so?
First, invert the fraction to get rid of the negative power outside the bracket: $\displaystyle \left(\frac{2x^{-2} y^3}{6xz^5}\right)^{-4}=\left(\frac{6xz^5}{2x^{-2} y^3}\right)^{4}$
Then move the $\displaystyle x^{-2}$ term to the numerator to get rid of its negative power:$\displaystyle =\left(\frac{6xz^5x^2}{2 y^3}\right)^{4}$
(By the way, what I should more correctly say is: multiply top-and-bottom by $\displaystyle x^2$, and then note that in the denominator $\displaystyle x^{-2}\times x^2 = 1$. But it's quicker to say: move the $\displaystyle x^{-2}$ term ...)
Simplify:$\displaystyle =\left(\frac{3x^3z^5}{y^3}\right)^{4}$
Lastly, remove the brackets by raising everything to the power $\displaystyle 4$:
$\displaystyle =\frac{81x^{12}z^{20}}{y^{12}}$
Grandad