# Question #13eae

May 19, 2016

${\left(\frac{{x}^{3} {y}^{4}}{{x}^{- 2} {y}^{- 4}}\right)}^{-} 1 = \frac{1}{{x}^{5} {y}^{8}} = {x}^{-} 5 {y}^{-} 8$

#### Explanation:

We will use the following properties of exponents:

• ${x}^{a} \cdot {x}^{b} = {x}^{a + b}$

• ${\left({x}^{a}\right)}^{b} = {x}^{a b}$

• ${\left(x y\right)}^{a} = {x}^{a} {y}^{a}$

• ${x}^{0} = 1$ for all $x \ne 0$

With those, for $x , y \ne 0$ we have

${\left(\frac{{x}^{3} {y}^{4}}{{x}^{- 2} {y}^{- 4}}\right)}^{-} 1 = \frac{{\left({x}^{3}\right)}^{- 1} {\left({y}^{4}\right)}^{- 1}}{{\left({x}^{- 2}\right)}^{- 1} {\left({y}^{- 4}\right)}^{- 1}}$

$= \frac{{x}^{3 \cdot - 1} {y}^{4 \cdot - 1}}{{x}^{- 2 \cdot - 1} {y}^{- 4 \cdot - 1}}$

$= \frac{{x}^{-} 3 {y}^{-} 4}{{x}^{2} {y}^{4}}$

$= \frac{\left({x}^{-} 3 {y}^{-} 4\right) \left({x}^{3} {y}^{4}\right)}{\left({x}^{2} {y}^{4}\right) \left({x}^{3} {y}^{4}\right)}$

$= \frac{{x}^{- 3 + 3} {y}^{- 4 + 4}}{{x}^{2 + 3} {y}^{4 + 4}}$

$= \frac{{x}^{0} {y}^{0}}{{x}^{5} {y}^{8}}$

$= \frac{1 \cdot 1}{{x}^{5} {y}^{8}}$

$= \frac{1}{{x}^{5} {y}^{8}}$

Note that we can also multiply this by $\frac{{x}^{-} 5 {y}^{-} 8}{{x}^{-} 5 {y}^{-} 8}$ to find it to be equivalent to ${x}^{-} 5 {y}^{-} 8$