# How do you solve 3(x+1)^(4/3)=48 ?

May 7, 2016

$x = 7$

#### Explanation:

$3 {\left(x + 1\right)}^{\frac{4}{3}} = 48$

Divide both sides by $\textcolor{red}{3}$

$3 {\left(x + 1\right)}^{\frac{4}{3}} / \textcolor{red}{3} = \frac{48}{\textcolor{red}{3}}$

${\left(x + 1\right)}^{\frac{4}{3}} = 16$

We know that ${2}^{4} = 16$

$\implies {\left(x + 1\right)}^{\frac{4}{3}} = {2}^{4}$

Find the 4th root of the equation:

$\implies {\left(x + 1\right)}^{\frac{4}{3} \times \textcolor{red}{\frac{1}{4}}} = {2}^{4 \times \textcolor{red}{\frac{1}{4}}}$

$\implies {\left(x + 1\right)}^{\frac{1}{3}} = 2$

Now, we find the cube of the equation:

$\implies {\left(x + 1\right)}^{\frac{\textcolor{red}{3}}{3}} = {2}^{\textcolor{red}{3}}$

$\implies x + 1 = {2}^{3}$

$\implies x + 1 = 8$

Subtract 1 from both sides

$\implies x + 1 \textcolor{red}{- 1} = 8 \textcolor{red}{- 1}$

$\implies x = 7$

$3 {\left(x + 1\right)}^{\frac{4}{3}}$

$= 3 {\left(7 + 1\right)}^{\frac{4}{3}}$

$= 3 {\left(8\right)}^{\frac{4}{3}}$

$= 3 \times {2}^{4}$ because, $\textcolor{red}{{8}^{\frac{1}{3}} = \sqrt[3]{8} = 2}$

$= 3 \times 16$

$= 48$

Therefore, $x = 7$