How do you simplify #((3x^5y^4)/(x^0y^-3))^2# and write it using only positive exponents?

1 Answer
Feb 16, 2017

Answer:

See the entire simplification process below:

Explanation:

First, I would use this rule of exponents to simplify the denominator of the fraction: #a^color(red)(0) = 1#

#((3x^5y^4)/(x^color(red)(0)y^-3))^2 = ((3x^5y^4)/(1y^-3))^2 = ((3x^5y^4)/y^-3)^2#

Next, I would use this rule of exponents to simplify the terms within parenthesis: #x^color(red)(a)/x^color(blue)(b) = x^(color(red)(a)-color(blue)(b))#

#((3x^5y^color(red)(4))/y^color(blue)(-3))^2 = (3x^5y^(color(red)(4)-color(blue)(-3)))^2 = (3x^5y^7)^2#

Now, I would use these two rules of exponents to complete the simplification: #a = a^color(red)(1)# and #(x^color(red)(a))^color(blue)(b) = x^(color(red)(a) xx color(blue)(b))#

#(3x^5y^7)^2 = (3^color(red)(1)x^color(red)(5)y^color(red)(7))^color(blue)(2) = 3^(color(red)(1) xx color(blue)(2))x^(color(red)(5)xxcolor(blue)(2))y^(color(red)(7)xxcolor(blue)(2)) = 3^2x^10y^14 = 9x^10y^14#