Solve for k (3 square root of 3)^2 = square root of k

${(3 \sqrt{3})}^{2} = \sqrt{k}$

Simplify ${(3 \sqrt{3})}^{2}$ .

Simplify the expression.

Apply the product rule to $3 \sqrt{3}$ .

$3^{2} {\sqrt{3}}^{2} = \sqrt{k}$

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

$9 {\sqrt{3}}^{2} = \sqrt{k}$

Rewrite ${\sqrt{3}}^{2}$ as $3$ .

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

$9 {(3^{\frac{1}{2}})}^{2} = \sqrt{k}$

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

$9 \cdot 3^{\frac{1}{2} \cdot 2} = \sqrt{k}$

Combine $\frac{1}{2}$ and $2$ .

$9 \cdot 3^{\frac{2}{2}} = \sqrt{k}$

Cancel the common factor of $2$ .

Cancel the common factor.

$9 \cdot 3^{\frac{2}{2}} = \sqrt{k}$

Rewrite the expression.

$9 \cdot 3^{1} = \sqrt{k}$

Evaluate the exponent.

$9 \cdot 3 = \sqrt{k}$

Multiply $9$ by $3$ .

$27 = \sqrt{k}$

Rewrite the equation as $\sqrt{k} = 27$ .

$\sqrt{k} = 27$

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

${\sqrt{k}}^{2} = 27^{2}$

Simplify each side of the equation.

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

${(k^{\frac{1}{2}})}^{2} = 27^{2}$

Simplify the left side.

Simplify ${(k^{\frac{1}{2}})}^{2}$ .

Multiply the exponents in ${(k^{\frac{1}{2}})}^{2}$ .

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

$k^{\frac{1}{2} \cdot 2} = 27^{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

$k^{\frac{1}{2} \cdot 2} = 27^{2}$

Rewrite the expression.

$k^{1} = 27^{2}$

Simplify.

$k = 27^{2}$

Simplify the right side.

Raise $27$ to the power of $2$ .

$k = 729$