Solve for n (-1-n)^(4/5)=(3n+11)^(4/5)

${(- 1 - n)}^{\frac{4}{5}} = {(3 n + 11)}^{\frac{4}{5}}$

Step 1

Since the exponents are equal, the bases of the exponents on both sides of the equation must be equal.

$- 1 - n = 3 n + 11$

Step 2

Solve for $n$ .

Move all terms containing $n$ to the left side of the equation.

Subtract $3 n$ from both sides of the equation.

$- 1 - n - 3 n = 11$

Subtract $3 n$ from $- n$ .

$- 1 - 4 n = 11$

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

Add $1$ to both sides of the equation.

$- 4 n = 11 + 1$

Add $11$ and $1$ .

$- 4 n = 12$

Divide each term in $- 4 n = 12$ by $- 4$ and simplify.

Divide each term in $- 4 n = 12$ by $- 4$ .

$\frac{- 4 n}{- 4} = \frac{12}{- 4}$

Simplify the left side.

Cancel the common factor of $- 4$ .

Cancel the common factor.

$\frac{- 4 n}{- 4} = \frac{12}{- 4}$

Divide $n$ by $1$ .

$n = \frac{12}{- 4}$

Simplify the right side.

Divide $12$ by $- 4$ .

$n = - 3$