How do you evaluate and simplify #64^(-2/3)#?

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Answer 1 · 2017-06-30T13:29:48

See a solution process below:

First, rewrite the expression as:

#64^((1/3 xx -2))#

Using this rule of exponents we can rewrite this as:

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

#64^(color(red)(1/3) xx color(blue)(-2)) = (64^color(red)(1/3))^color(blue)(-2)#

Next, use this rule to again rewrite and evaluate the term within the parenthesis:

#x^(1/color(red)(n)) = root(color(red)(n))(x)#

#(64^(1/color(red)(n)))^-2 = (root(color(red)(3))(64))^-2 = 4^-2#

Now, use this rule to complete the evaluation:

#4^color(red)(-2) = 1/4^color(red)(- -2) = 1/4^2 = 1/16#

Answer 2 · 2017-07-09T17:48:19

#= 1/16#

The laws of indices state.

#x^-m = 1/x^m#

#x^(p/q) = (rootq x)^p#

Using these laws gives the following:

#64^(-2/3) = 1/64^(2/3)" "larr# make the index positive

#1/64^(2/3) = 1/root3 64^2" "larr# find the cube root

#=1/(4^2)" "larr# find the square

#= 1/16#