How do you simplify #(x^(-1/2)y^4)^(1/4)/(x^(2/3)y^(3/2)*x^(-3/2)y^(1/2))#?

1 Answer
Apr 11, 2018

#(x^(17/24))/(y)#

Explanation:

Simplify: #(x^(-1/2)y^4)^(1/4)/(x^(2/3)y^(3/2)x^(-3/2)y^(1/2))#

First, let's distribute the #1/4# exponent:

#(x^((-1/2)^(1/4))y^((4)^(1/4)))/(x^(2/3)y^(3/2)x^(-3/2)y^(1/2))#

When you raise a power to a power you multiply them to find the exponent. Let's do that

#(x^((-1/2)(1/4))y^((4)(1/4)))/(x^(2/3)y^(3/2)x^(-3/2)y^(1/2))#

#(x^(-1/8)y^1)/(x^(2/3)y^(3/2)x^(-3/2)y^(1/2))#

#(x^(-1/8)y)/(x^(2/3)y^(3/2)x^(-3/2)y^(1/2))#

Now let's combine the exponents with like bases in the denominator. When you multiply exponents with like bases you add the exponents together

#(x^(-1/8)y)/(x^((2/3)+(-3/2))y^((3/2)+(1/2)))#

#(x^(-1/8)y)/(x^((4/6)+(-9/6))y^(4/2))#

#(x^(-1/8)y)/(x^(-5/6)y^(2))#

When there is a negative exponent in the numerator we can bring it to the denominator as a positive exponent and vice versa

Let's bring #x^(-5/6)# to the numerator since it is bigger than #x^(-1/8)# so when we multiply them and add the exponents it will be positive.

#(x^(-1/8)x^(5/6)y)/(y^(2))#

#(x^((-1/8)+(5/6))y)/(y^(2))#

#(x^((-3/24)+(20/24))y)/(y^(2))#

#(x^(17/24)y)/(y^(2))#

Now we can cancel out one of the #y#'s on top and bottom

#(x^(17/24)cancel(y))/(y^(cancel(2)color(red)(1)))#

#(x^(17/24))/(y)#

The final answer is

#(x^(17/24))/(y)#