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

1 Answer
Feb 15, 2017

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)