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

1 Answer
Jun 20, 2016

Answer:

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.