How do you simplify #(4n ) ^ { - 9/ 7}#?

1 Answer
Parabola
Dec 9, 2017

#(root[7] (4^5xxn^5))/(16n^2)#

Explanation:

Remember this rule:
#a^(-x/y)=1/(root[y] (a^x))#

Therefore,
#(4n)^(-9/7)=1/(root[7] ((4n)^9)#

We simplify the root.
#1/(root[7] ((4n)^9)# becomes
#1/(root[7] ((4^9xxn^9))# We can take #4^7# and #n^7# from the root.
#1/(root[7] ((4^9xxn^9))# =>#1/(4nroot[7] (4^2xxn^2))#
We simplify the denominator.
#(1xx4^(5/7)xxn^(5/7))/(4nroot[7] (4^2xxn^2)xx4^(5/7)xxn^(5/7))# =>#(4^(5/7)xxn^(5/7))/(4xx4xxnxxn)# which is really
#(root[7] (4^5xxn^5))/(16n^2)#

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