# How do you solve -(x-5)^(1/4)+7/3=2?

May 24, 2017

$x = \frac{406}{81}$

#### Explanation:

Subtract $\frac{7}{3}$ from both sides.

$- {\left(x - 5\right)}^{\frac{1}{4}} + \frac{7}{3} - \textcolor{red}{\frac{7}{3}} = 2 - \textcolor{red}{\frac{7}{3}}$

$\textcolor{w h i t e}{\text{XXXXX-.}} - {\left(x - 5\right)}^{\frac{1}{4}} = - \frac{1}{3}$

Now multiply both sides by $- 1$.

$\textcolor{b l u e}{- 1} \left(- {\left(x - 5\right)}^{\frac{1}{4}}\right) = \textcolor{b l u e}{- 1} \left(- \frac{1}{3}\right)$

$\textcolor{w h i t e}{\text{XXXXX}} {\left(x - 5\right)}^{\frac{1}{4}} = \frac{1}{3}$

Next, take both sides to the power of $4$.

${\textcolor{\mathmr{and} a n \ge}{\left({\textcolor{b l a c k}{\left(x - 5\right)}}^{\textcolor{b l a c k}{\frac{1}{4}}}\right)}}^{\textcolor{\mathmr{and} a n \ge}{4}} = {\textcolor{\mathmr{and} a n \ge}{\left(\textcolor{b l a c k}{\frac{1}{3}}\right)}}^{\textcolor{\mathmr{and} a n \ge}{4}}$

$\left(x - 5\right) = \frac{1}{81}$

Finally, add $5$ to both sides.

$x - 5 + \textcolor{\lim e g r e e n}{5} = \frac{1}{81} + \textcolor{\lim e g r e e n}{5}$

$x = \frac{406}{81}$