Question

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

Answer 1

Simplified, your expression is equivalent to `{x^4y}/7`

Explanation:

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

`b^-n=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.

`10/color(red)(5)=2,10/2=color(red)(5)`
So in short: `x=1/{1/x}`.

Applying the first demonstrated identity will look like so:

`1/{7x^-4y^-1} rArr 1/{7{1}/{x^4}{1}/{y}} rArr 1/{7{1}/{x^4y}}`

Now we will apply the second demonstrated rule:

`1/{7{1}/{x^4y}} rArr {x^4y}/7`

Yielding the simplified expression.