Question

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`

Answer 1

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`