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

1 Answer
Apr 13, 2018

Answer:

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 #(x^5y^-4)^-2# into just exponents of x and y.

Since #(a^b)^c=a^(bc)#, this can simplify into #x^-10y^8#, so the whole equation becomes #x^-2/(x^-10y^8)#.

Additionally, since #a^-b=1/a^b#, you can turn the #x^-2# in the numerator into #1/x^2#, and the #x^-10# in the denominator into #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 #1/a^b# values:

#1/x^2÷(1/x^10y^8)# can also be written as #1/x^2*(x^10 1/y^8)# (just like when you divide fractions).

Therefore, the equation can now be written as #x^10/(x^2y^8)#. 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!