How do you solve ${ln (x + 2)}^{3} = 4$ $ln(x+2)^3=4$ ?

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Answer 1 · 2016-07-18T15:21:41

$ln (x + 2) = \sqrt[3]{4}$ $ln(x+2)= root(3)4$

$x + 2 = e^{\sqrt[3]{4}}$ $x+2= e^(root(3)4)$

$x = e^{\sqrt[3]{4}} - 2$ $x= e^(root(3)4)-2$

How do you solve ln(x+2)3=4ln(x+2)^3=4?