# How do you simplify (2x^-3y^-2)/z^4 using only positive exponents?

$\frac{2}{{x}^{3} {y}^{2} {z}^{4}}$
Since ${x}^{- 1} = \frac{1}{x}$, we have ${x}^{- 3} = \frac{1}{x} ^ 3$ and ${y}^{- 2} = \frac{1}{y} ^ 2$, so you get
$\frac{2 {x}^{- 3} {y}^{- 2}}{z} ^ 4 = \frac{2}{{x}^{3} {y}^{2} {z}^{4}}$