Solve for x cube root of 2x+1=2

$\sqrt[3]{2 x + 1} = 2$

Step 1

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

${\sqrt[3]{2 x + 1}}^{3} = 2^{3}$

Step 2

Simplify each side of the equation.

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

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

Simplify the left side.

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

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

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

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

Cancel the common factor of $3$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify.

$2 x + 1 = 2^{3}$

Simplify the right side.

Raise $2$ to the power of $3$ .

$2 x + 1 = 8$

Step 3

Solve for $x$ .

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

Subtract $1$ from both sides of the equation.

$2 x = 8 - 1$

Subtract $1$ from $8$ .

$2 x = 7$

Divide each term in $2 x = 7$ by $2$ and simplify.

Divide each term in $2 x = 7$ by $2$ .

$\frac{2 x}{2} = \frac{7}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x}{2} = \frac{7}{2}$

Divide $x$ by $1$ .

$x = \frac{7}{2}$

Step 4

The result can be shown in multiple forms.

Exact Form:

$x = \frac{7}{2}$

Decimal Form:

$x = 3.5$

Mixed Number Form:

$x = 3 \frac{1}{2}$

Solve for x cube root of 2x+1=2

Related Questions

Support LCMGCF.com by disabling your adblocker.