# How do you simplify x^-2/(x^5y^-4)^-2 and write it using only positive exponents?

Apr 13, 2018

The answer is ${x}^{8} / {y}^{8}$.

#### Explanation:

Note: when the variables $a$, $b$, and $c$ are used, I am referring to a general rule that will work for every real value of $a$, $b$, or $c$.

First, you have to look at the denominator and expand out ${\left({x}^{5} {y}^{-} 4\right)}^{-} 2$ into just exponents of x and y.

Since ${\left({a}^{b}\right)}^{c} = {a}^{b c}$, this can simplify into ${x}^{-} 10 {y}^{8}$, so the whole equation becomes ${x}^{-} \frac{2}{{x}^{-} 10 {y}^{8}}$.

Additionally, since ${a}^{-} b = \frac{1}{a} ^ b$, you can turn the ${x}^{-} 2$ in the numerator into $\frac{1}{x} ^ 2$, and the ${x}^{-} 10$ in the denominator into $\frac{1}{x} ^ 10$.

Therefore, the equation can be rewritten as such:
(1/x^2)/((1/x^10y^8). However, to simplify this, we need to get rid of the $\frac{1}{a} ^ b$ values:

1/x^2÷(1/x^10y^8) can also be written as $\frac{1}{x} ^ 2 \cdot \left({x}^{10} \frac{1}{y} ^ 8\right)$ (just like when you divide fractions).

Therefore, the equation can now be written as ${x}^{10} / \left({x}^{2} {y}^{8}\right)$. However, there are $x$ values on both the numerator and the denominator.

Since a^b/a^c=a^(b-c , you can simplify this as ${x}^{8} / {y}^{8}$.

Hope this helps!