How do you solve #(n-27)^(3/2) = 64#?

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Answer 1 · 2017-05-05T21:08:38

#n=43#

Proceed as follows

Raise both sides of the equation to the #2/3# power

#[(n-27)^(3/2)]^(2/3)=(64)^(2/3)#

Using properties of exponents rewrite

#n-27=[64^(1/3)]^2#

The cube root of #64# is #4#

#n-27=(4)^2#

#n-27=16#

#n=27+16=43#