Solve for x fifth root of 7x+3-8=-10

$\sqrt[5]{7 x + 3} - 8 = - 10$

Move all terms not containing $\sqrt[5]{7 x + 3}$ to the right side of the equation.

Add $8$ to both sides of the equation.

$\sqrt[5]{7 x + 3} = - 10 + 8$

Add $- 10$ and $8$ .

$\sqrt[5]{7 x + 3} = - 2$

To remove the radical on the left side of the equation, raise both sides of the equation to the power of $5$ .

${\sqrt[5]{7 x + 3}}^{5} = {(- 2)}^{5}$

Simplify each side of the equation.

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt[5]{7 x + 3}$ as ${(7 x + 3)}^{\frac{1}{5}}$ .

${({(7 x + 3)}^{\frac{1}{5}})}^{5} = {(- 2)}^{5}$

Simplify the left side.

Simplify ${({(7 x + 3)}^{\frac{1}{5}})}^{5}$ .

Multiply the exponents in ${({(7 x + 3)}^{\frac{1}{5}})}^{5}$ .

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

${(7 x + 3)}^{\frac{1}{5} \cdot 5} = {(- 2)}^{5}$

Cancel the common factor of $5$ .

Cancel the common factor.

${(7 x + 3)}^{\frac{1}{5} \cdot 5} = {(- 2)}^{5}$

Rewrite the expression.

${(7 x + 3)}^{1} = {(- 2)}^{5}$

Simplify.

$7 x + 3 = {(- 2)}^{5}$

Simplify the right side.

Raise $- 2$ to the power of $5$ .

$7 x + 3 = - 32$

Solve for $x$ .

Move all terms not containing $x$ to the right side of the equation.

Subtract $3$ from both sides of the equation.

$7 x = - 32 - 3$

Subtract $3$ from $- 32$ .

$7 x = - 35$

Divide each term in $7 x = - 35$ by $7$ and simplify.

Divide each term in $7 x = - 35$ by $7$ .

$\frac{7 x}{7} = \frac{- 35}{7}$

Simplify the left side.

Cancel the common factor of $7$ .

Cancel the common factor.

$\frac{7 x}{7} = \frac{- 35}{7}$

Divide $x$ by $1$ .

$x = \frac{- 35}{7}$

Simplify the right side.

Divide $- 35$ by $7$ .

$x = - 5$