Precalculus Examples

${(\sqrt{k + 7})}^{2} = {(3 \sqrt{2})}^{2}$

Step 1

Rewrite ${\sqrt{k + 7}}^{2}$ as $k + 7$ .

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Step 1.1

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

${({(k + 7)}^{\frac{1}{2}})}^{2} = {(3 \sqrt{2})}^{2}$

Step 1.2

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

${(k + 7)}^{\frac{1}{2} \cdot 2} = {(3 \sqrt{2})}^{2}$

Step 1.3

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

${(k + 7)}^{\frac{2}{2}} = {(3 \sqrt{2})}^{2}$

Step 1.4

Cancel the common factor of $2$ .

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Step 1.4.1

Cancel the common factor.

${(k + 7)}^{\frac{2}{2}} = {(3 \sqrt{2})}^{2}$

Step 1.4.2

Rewrite the expression.

${(k + 7)}^{1} = {(3 \sqrt{2})}^{2}$

Step 1.5

Simplify.

$k + 7 = {(3 \sqrt{2})}^{2}$

Step 2

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

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Step 2.1

Simplify the expression.

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Step 2.1.1

Apply the product rule to $3 \sqrt{2}$ .

$k + 7 = 3^{2} {\sqrt{2}}^{2}$

Step 2.1.2

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

$k + 7 = 9 {\sqrt{2}}^{2}$

Step 2.2

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

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Step 2.2.1

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

$k + 7 = 9 {(2^{\frac{1}{2}})}^{2}$

Step 2.2.2

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

$k + 7 = 9 \cdot 2^{\frac{1}{2} \cdot 2}$

Step 2.2.3

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

$k + 7 = 9 \cdot 2^{\frac{2}{2}}$

Step 2.2.4

Cancel the common factor of $2$ .

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Step 2.2.4.1

Cancel the common factor.

$k + 7 = 9 \cdot 2^{\frac{2}{2}}$

Step 2.2.4.2

Rewrite the expression.

$k + 7 = 9 \cdot 2^{1}$

Step 2.2.5

Evaluate the exponent.

$k + 7 = 9 \cdot 2$

Step 2.3

Multiply $9$ by $2$ .

$k + 7 = 18$

Step 3

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

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Step 3.1

Subtract $7$ from both sides of the equation.

$k = 18 - 7$

Step 3.2

Subtract $7$ from $18$ .

$k = 11$