Solve for x 12+ cube root of 2x-7=15

$12 + \sqrt[3]{2 x - 7} = 15$

Move all terms not containing $\sqrt[3]{2 x - 7}$ to the right side of the equation.

Subtract $12$ from both sides of the equation.

$\sqrt[3]{2 x - 7} = 15 - 12$

Subtract $12$ from $15$ .

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

To remove the radical on the left side of the equation, cube both sides of the equation.

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

Simplify each side of the equation.

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt[3]{2 x - 7}$ as ${(2 x - 7)}^{\frac{1}{3}}$ .

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

Simplify the left side.

Simplify ${({(2 x - 7)}^{\frac{1}{3}})}^{3}$ .

Multiply the exponents in ${({(2 x - 7)}^{\frac{1}{3}})}^{3}$ .

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

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

Cancel the common factor of $3$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify.

$2 x - 7 = 3^{3}$

Simplify the right side.

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

$2 x - 7 = 27$

Solve for $x$ .

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

Add $7$ to both sides of the equation.

$2 x = 27 + 7$

Add $27$ and $7$ .

$2 x = 34$

Divide each term in $2 x = 34$ by $2$ and simplify.

Divide each term in $2 x = 34$ by $2$ .

$\frac{2 x}{2} = \frac{34}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x}{2} = \frac{34}{2}$

Divide $x$ by $1$ .

$x = \frac{34}{2}$

Simplify the right side.

Divide $34$ by $2$ .

$x = 17$