How do you simplify #(64n^12)^(-1/6)#?

Question

4644 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-01-16T23:40:12

See the entire simplification process below:

First, we will use these rule of exponents to start the simplification process:

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

#a = a^color(red)(1)#

#(64n^color(red)(12))^(color(blue)(-1/6)) = (64^color(red)(1)n^color(red)(12))^(color(blue)(-1/6)) =#

#64^(color(red)(1) xx color(blue)(-1/6))n^(color(red)(12) xx color(blue)(-1/6)) = 64^(-1/6)n^-2#

Now we can use this rule for exponents to continue the simplification process:

#x^color(red)(a) = 1/x^color(red)(-a)#

#64^(-1/6)n^-2 = 1/(64^(- -1/6)n^(- -2)) = 1/(64^(1/6)n^2)#

Now, to finish we need to know #2^6 = 64# and #(-2)^6 = 64# therefore #64^(1/6) = 2# or #-2#

#1/(64^(1/6)n^2) = 1/(2n^2)# or #1/(64^(1/6)n^2) = -1/(2n^2)#