# How do you simplify (4x^4y^-3)^-2 and write it using only positive exponents?

Jan 8, 2017

#### Answer:

See full simplification process below

#### Explanation:

First, we we use this rule for exponents to start the simplification:

${\left({x}^{\textcolor{red}{a}}\right)}^{\textcolor{b l u e}{b}} = {x}^{\textcolor{red}{a} \times \textcolor{b l u e}{b}}$

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

${4}^{-} 2 {x}^{4 \times - 2} {y}^{- 3 \times - 2} \to$

${4}^{-} 2 {x}^{-} 8 {y}^{6}$

Now, we will use this rule for exponents to finalize the simplification leaving only terms with positive exponents:

${x}^{\textcolor{red}{a}} = \frac{1}{x} ^ \textcolor{red}{- a}$

y^6/(4^(- -2)x^(- -8)

y^6/(4^2x^8

${y}^{6} / \left(16 {x}^{8}\right)$

Jan 8, 2017

#### Answer:

$\frac{{y}^{6}}{16 {x}^{8}}$

#### Explanation:

Using the $\textcolor{b l u e}{\text{laws of exponents}}$

$\textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{2}{2}} \textcolor{b l a c k}{{\left({a}^{m} {b}^{n}\right)}^{p} = {a}^{m p} {b}^{n p}} \textcolor{w h i t e}{\frac{2}{2}} |}}}$
This can be extended to more than 2 products.

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

$= {4}^{- 2} {x}^{- 8} {y}^{6}$

$\textcolor{\mathmr{and} a n \ge}{\text{Reminder }} \textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{2}{2}} \textcolor{b l a c k}{{a}^{-} m \Leftrightarrow \frac{1}{a} ^ m} \textcolor{w h i t e}{\frac{2}{2}} |}}}$

$\Rightarrow {4}^{- 2} {x}^{- 8} {y}^{6} = \frac{1}{4} ^ 2 \times \frac{1}{x} ^ 8 \times {y}^{6} / 1$

$= \frac{1 \times 1 \times {y}^{6}}{16 \times {x}^{8} \times 1} = \frac{{y}^{6}}{16 {x}^{8}}$