How do you simplify #(ab)^-1/(cd^-2)# and write it using only positive exponents?

1 Answer
Feb 15, 2017

Answer:

See the entire simplification process below:

Explanation:

First, let's simplify the numerator using these two rules for exponents:

#a = a^color(red)(1)# and #(x^color(red)(a))^color(blue)(b) = x^(color(red)(a) xx color(blue)(b))#

#(ab)^-1/(cd^-2) = (a^color(red)(1)b^color(red)(1))^color(blue)(-1)/(cd^-2) = (a^(color(red)(1)xxcolor(blue)(-1))b^(color(red)(1)xxcolor(blue)(-1)))/(cd^-2) = (a^-1b^-1)/(cd^-2)#

Now, we can deal with the terms with negative exponents using these rules for exponents:

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

#(a^color(red)(-1)b^color(red)(-1))/(cd^color(red)(-2)) = d^color(red)(- -2)/(a^color(red)(- -1)b^color(red)(- -1)c) = d^color(red)(- -2)/(a^color(red)(- -1)b^color(red)(- -1)c) = d^color(red)(2)/(a^color(red)(1)b^color(red)(1)c)#

Now, we can finalize the simplification by applying this rule for exponents:

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

#d^color(red)(2)/(a^color(red)(1)b^color(red)(1)c) = d^2/(abc)#