How do you simplify #(2x^4y^-3)^-1# and write it using only positive exponents?

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Answer 1 · 2017-05-17T22:15:48

See a solution process below:

First, eliminate the outer negative exponent using this rule for exponents:

#x^color(red)(a) = 1/x^color(red)(-a)#

#(2x^4y^-3)^color(red)(-1) => 1/(2x^4y^-3)^color(red)(- -1) => 1/(2x^4y^-3)^color(red)(1)#

Next, use this rule of exponents to eliminate the outer exponent:

#a^color(red)(1) = a#

#1/(2x^4y^-3)^color(red)(1) => 1/(2x^4y^-3)#

Now, use this rule of exponents to eliminate the negative exponent for the #y# term:

#1/x^color(red)(a) = x^color(red)(-a)#

#1/(2x^4y^color(red)(-3)) => y^color(red)(- -3)/(2x^4) => y^3/(2x^4)#

Answer 2 · 2017-05-17T23:00:26

#y^3/(2x^4#

Smendyka has shown a method whereby the negative index outside the bracket is changed to a positive in the denominator.

Here is another method:

Recall the laws of indices:

#(x^3)^4 = x^(3xx4) = x^12 and (xyz)^m = x^my^mz^m#

#(2x^4y^-3)^-1" "larr# multiply the indices

# = color(blue)(2^-1x^-4)y^3" "larr# move to the denominator

#=y^3/(color(blue)(2^1x^4)#