How do you simplify [(4x^-4 y^2)/ (5x^6 y^-3)]^-3 leaving only positive exponents?

1 Answer
May 16, 2016

(125x^30)/(64y^15)

Explanation:

Given,

((4x^-4y^2)/(5x^6y^-3))^color(blue)(-3)

According to the exponent law , (xy)^n=x^ny^n, the color(blue)(-3) multiplies by each exponent in the numerator and denominator. In addition, the numbers are raised to color(blue)(-3). Thus,

=(4^(color(blue)(-3))x^((-4xxcolor(blue)(-3)))y^((2xxcolor(blue)(-3))))/(5^(color(blue)(-3))x^((6xxcolor(blue)(-3)))y^((-3xxcolor(blue)(-3))))

Simplifying,

=(4^-3x^12y^-6)/(5^-3x^-18y^9)

According to the exponent law , x^-n/1=1/x^n. Similarly, 1/x^-n=x^n/1. Thus,

=(5^3x^(12)*5x^18)/(4^3y^9y^6)

Simplifying,

=(125x^(12)x^18)/(64y^9y^6)

Using the exponent law , x^mx^n=x^(m+n), the exponents in the numerator of base x can be added together. Similarly, the exponents in the denominator of base y can be added together.

=(125x^(12+18))/(64y^(9+6))

=color(green)(|bar(ul(color(white)(a/a)color(black)((125x^30)/(64y^15))color(white)(a/a)|)))