Solve for r, s, and t?

(\frac{x^4 y^3 z^2 x^{-5}}{x^5 y^2 z^2 y^4})^{-3} equals x^ry^sz^t

1 Answer
Feb 20, 2018

See a solution process below:

Explanation:

First, use this rule of exponents to combine the x terms in the numerator and y terms in the denominator:

x^color(red)(a) xx x^color(blue)(b) = x^(color(red)(a) + color(blue)(b))

((x^color(red)(4)y^3z^2x^color(blue)(-5))/(x^5y^color(red)(2)z^2y^color(blue)(4)))^-3 =>

((x^color(red)(4)x^color(blue)(-5)y^3z^2)/(x^5y^color(red)(2)y^color(blue)(4)z^2))^-3 =>

((x^(color(red)(4)+color(blue)(-5))y^3z^2)/(x^5y^(color(red)(2)+color(blue)(4))z^2))^-3 =>

((x^(color(red)(4)-color(blue)(5))y^3z^2)/(x^5y^6z^2))^-3 =>

((x^-1y^3z^2)/(x^5y^6z^2))^-3

Next, use this rule of exponents to combine the common terms:

x^color(red)(a)/x^color(blue)(b) = x^(color(red)(a)-color(blue)(b))

((x^color(red)(-1)y^color(red)(3)z^color(red)(2))/(x^color(blue)(5)y^color(blue)(6)z^color(blue)(2)))^-3 =>

(x^(color(red)(-1)-color(blue)(5))y^(color(red)(3)-color(blue)(6))z^(color(red)(2)-color(blue)(2)))^-3 =>

(x^-6y^-3z^0)^-3

Now, use this rule of exponents to complete the simplification:

(x^color(red)(a))^color(blue)(b) = x^(color(red)(a) xx color(blue)(b))

x^(color(red)(-6) xx color(blue)(-3))y^(color(red)(-3) xx color(blue)(-3))z^(color(red)(0) xx color(blue)(-3)) =>

x^18y^-9z^0

Therefore:

x^r = x^18 => r = 18

y^s = y^-9 => s = -9

z^t = z^0 => t = 0