How do you write #(9y^6) ^(3/2)# as an radical form?

1 Answer
Y
May 10, 2017

Numerator is what we take the power to, denominator is which root we take the base of. Answer: #sqrt((9y^6)^3)#

Explanation:

Formatted problem: #(9y^6)^(3/2)#

When we have a fractional exponent, the number in the numerator is what we take the base to the power of and the number in the denominator is what root we take of the base.
For example #a^(x/y)#, where #y!=0#, would be equal to #rooty(a^x)# or #rooty(a)^x#.

Using this concept, we can write #(9y^6)^(3/2)# as:
#sqrt((9y^6)^3)# which is the given question in radical form.

If necessary for the problem, we can simplify:
#=sqrt((3^2y^6))^3#
#=(3y^3)^3#
#=27y^9#

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