Solve for n fourth root of n+2=3

$\sqrt[4]{n + 2} = 3$

To remove the radical on the left side of the equation, raise both sides of the equation to the power of $4$ .

${\sqrt[4]{n + 2}}^{4} = 3^{4}$

Simplify each side of the equation.

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt[4]{n + 2}$ as ${(n + 2)}^{\frac{1}{4}}$ .

${({(n + 2)}^{\frac{1}{4}})}^{4} = 3^{4}$

Simplify the left side.

Simplify ${({(n + 2)}^{\frac{1}{4}})}^{4}$ .

Multiply the exponents in ${({(n + 2)}^{\frac{1}{4}})}^{4}$ .

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

${(n + 2)}^{\frac{1}{4} \cdot 4} = 3^{4}$

Cancel the common factor of $4$ .

Cancel the common factor.

${(n + 2)}^{\frac{1}{4} \cdot 4} = 3^{4}$

Rewrite the expression.

${(n + 2)}^{1} = 3^{4}$

Simplify.

$n + 2 = 3^{4}$

Simplify the right side.

Raise $3$ to the power of $4$ .

$n + 2 = 81$

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

Subtract $2$ from both sides of the equation.

$n = 81 - 2$

Subtract $2$ from $81$ .

$n = 79$