# How do you simplify 1/ (7x^-4y^-1) with positive exponents?

Jun 20, 2016

Simplified, your expression is equivalent to $\frac{{x}^{4} y}{7}$

#### Explanation:

All negative exponents with non-zero bases hold the following identity:

${b}^{-} n = \frac{1}{b} ^ n$

To be sure that this question can be sensibly solved, I will also elaborate that a reciprocal is a self-inverse function.

Much like subtraction (Example: (10-(10-color(red)(3)))=color(red)(3)), a reciprocal can be applied to itself to get the original value.

$\frac{10}{\textcolor{red}{5}} = 2 , \frac{10}{2} = \textcolor{red}{5}$
So in short: $x = \frac{1}{\frac{1}{x}}$.

Applying the first demonstrated identity will look like so:

$\frac{1}{7 {x}^{-} 4 {y}^{-} 1} \Rightarrow \frac{1}{7 \frac{1}{{x}^{4}} \frac{1}{y}} \Rightarrow \frac{1}{7 \frac{1}{{x}^{4} y}}$

Now we will apply the second demonstrated rule:

$\frac{1}{7 \frac{1}{{x}^{4} y}} \Rightarrow \frac{{x}^{4} y}{7}$

Yielding the simplified expression.