# How do you simplify (2x^4y^-3)^-1 and write it using only positive exponents?

May 17, 2017

See a solution process below:

#### Explanation:

First, eliminate the outer negative exponent using this rule for exponents:

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

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

Next, use this rule of exponents to eliminate the outer exponent:

${a}^{\textcolor{red}{1}} = a$

$\frac{1}{2 {x}^{4} {y}^{-} 3} ^ \textcolor{red}{1} \implies \frac{1}{2 {x}^{4} {y}^{-} 3}$

Now, use this rule of exponents to eliminate the negative exponent for the $y$ term:

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

$\frac{1}{2 {x}^{4} {y}^{\textcolor{red}{- 3}}} \implies {y}^{\textcolor{red}{- - 3}} / \left(2 {x}^{4}\right) \implies {y}^{3} / \left(2 {x}^{4}\right)$

May 17, 2017

y^3/(2x^4

#### Explanation:

Smendyka has shown a method whereby the negative index outside the bracket is changed to a positive in the denominator.

Here is another method:

Recall the laws of indices:

${\left({x}^{3}\right)}^{4} = {x}^{3 \times 4} = {x}^{12} \mathmr{and} {\left(x y z\right)}^{m} = {x}^{m} {y}^{m} {z}^{m}$

${\left(2 {x}^{4} {y}^{-} 3\right)}^{-} 1 \text{ } \leftarrow$ multiply the indices

$= \textcolor{b l u e}{{2}^{-} 1 {x}^{-} 4} {y}^{3} \text{ } \leftarrow$ move to the denominator

=y^3/(color(blue)(2^1x^4)