How do you solve ${(n - 27)}^{\frac{3}{2}} = 64$ $(n-27)^(3/2) = 64$ ?

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Answer 1 · 2017-05-05T21:08:38

$n = 43$ $n=43$

Proceed as follows

Raise both sides of the equation to the $\frac{2}{3}$ $2/3$ power

${[{(n - 27)}^{\frac{3}{2}}]}^{\frac{2}{3}} = {(64)}^{\frac{2}{3}}$ $[(n-27)^(3/2)]^(2/3)=(64)^(2/3)$

Using properties of exponents rewrite

$n - 27 = {[64^{\frac{1}{3}}]}^{2}$ $n-27=[64^(1/3)]^2$

The cube root of $64$ $64$ is $4$ $4$

$n - 27 = {(4)}^{2}$ $n-27=(4)^2$

$n - 27 = 16$ $n-27=16$

$n = 27 + 16 = 43$ $n=27+16=43$

How do you solve (n−27)32=64(n-27)^(3/2) = 64?