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 $4$

$n-27=(4)^2$

$n-27=16$

$n=27+16=43$

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