Solve for x fourth root of (x-3)^5=32

$\sqrt[4]{{(x - 3)}^{5}} = 32$

Step 1

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

${\sqrt[4]{{(x - 3)}^{5}}}^{4} = 32^{4}$

Step 2

Simplify each side of the equation.

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt[4]{{(x - 3)}^{5}}$ as ${(x - 3)}^{\frac{5}{4}}$ .

${({(x - 3)}^{\frac{5}{4}})}^{4} = 32^{4}$

Simplify the left side.

Multiply the exponents in ${({(x - 3)}^{\frac{5}{4}})}^{4}$ .

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

${(x - 3)}^{\frac{5}{4} \cdot 4} = 32^{4}$

Cancel the common factor of $4$ .

Cancel the common factor.

${(x - 3)}^{\frac{5}{4} \cdot 4} = 32^{4}$

Rewrite the expression.

${(x - 3)}^{5} = 32^{4}$

Simplify the right side.

Raise $32$ to the power of $4$ .

${(x - 3)}^{5} = 1048576$

Step 3

Solve for $x$ .

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$x - 3 = \sqrt[5]{1048576}$

Simplify $\sqrt[5]{1048576}$ .

Rewrite $1048576$ as $16^{5}$ .

$x - 3 = \sqrt[5]{16^{5}}$

Pull terms out from under the radical, assuming real numbers.

$x - 3 = 16$

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

Add $3$ to both sides of the equation.

$x = 16 + 3$

Add $16$ and $3$ .

$x = 19$