What is #(x^2y)^(1/2)#?

1 Answer
Jun 14, 2017

See a solution process below:

Explanation:

We can use these rules of exponents to simplify the expression:

#a = a^color(red)(1)# and #(x^color(red)(a))^color(blue)(b) = x^(color(red)(a) xx color(blue)(b))# and #a^color(red)(1) = a#

#(x^2y)^(1/2) => (x^color(red)(2)y^color(red)(1))^color(blue)(1/2) => x^(color(red)(2) xx color(blue)(1/2))y^(color(red)(1) xx color(blue)(1/2)) => x^color(red)(1)y^(1/2) => xy^(1/2)#

Or, if you want to write this in radical form:

#(x^2y)^(1/2) => sqrt(x^2y) => sqrt(x^2)sqrt(y) => xsqrt(y)#