Precalculus Examples

$\log (1) \sqrt{\frac{x^{2} + 4}{(x^{2} + 1) {(x^{3} - 7)}^{2}}}$

Step 1

Logarithm base $10$ of $1$ is $0$ .

$0 \sqrt{\frac{x^{2} + 4}{(x^{2} + 1) {(x^{3} - 7)}^{2}}}$

Step 2

Rewrite $\frac{x^{2} + 4}{(x^{2} + 1) {(x^{3} - 7)}^{2}}$ as ${(\frac{1}{x^{3} - 7})}^{2} \frac{x^{2} + 4}{x^{2} + 1^{2}}$ .

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Factor the perfect power $1^{2}$ out of $x^{2} + 4$ .

$0 \sqrt{\frac{1^{2} (x^{2} + 4)}{(x^{2} + 1) {(x^{3} - 7)}^{2}}}$

Factor the perfect power ${(x^{3} - 7)}^{2}$ out of $(x^{2} + 1) {(x^{3} - 7)}^{2}$ .

$0 \sqrt{\frac{1^{2} (x^{2} + 4)}{{(x^{3} - 7)}^{2} (x^{2} + 1^{2})}}$

Rearrange the fraction $\frac{1^{2} (x^{2} + 4)}{{(x^{3} - 7)}^{2} (x^{2} + 1^{2})}$ .

$0 \sqrt{{(\frac{1}{x^{3} - 7})}^{2} \frac{x^{2} + 4}{x^{2} + 1^{2}}}$

Step 3

Pull terms out from under the radical.

$0 (\frac{1}{x^{3} - 7} \sqrt{\frac{x^{2} + 4}{x^{2} + 1^{2}}})$

Step 4

One to any power is one.

$0 (\frac{1}{x^{3} - 7} \sqrt{\frac{x^{2} + 4}{x^{2} + 1}})$

Step 5

Rewrite $\sqrt{\frac{x^{2} + 4}{x^{2} + 1}}$ as $\frac{\sqrt{x^{2} + 4}}{\sqrt{x^{2} + 1}}$ .

$0 (\frac{1}{x^{3} - 7} \cdot \frac{\sqrt{x^{2} + 4}}{\sqrt{x^{2} + 1}})$

Step 6

Multiply $\frac{\sqrt{x^{2} + 4}}{\sqrt{x^{2} + 1}}$ by $\frac{\sqrt{x^{2} + 1}}{\sqrt{x^{2} + 1}}$ .

$0 (\frac{1}{x^{3} - 7} (\frac{\sqrt{x^{2} + 4}}{\sqrt{x^{2} + 1}} \cdot \frac{\sqrt{x^{2} + 1}}{\sqrt{x^{2} + 1}}))$

Step 7

Combine and simplify the denominator.

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Multiply $\frac{\sqrt{x^{2} + 4}}{\sqrt{x^{2} + 1}}$ by $\frac{\sqrt{x^{2} + 1}}{\sqrt{x^{2} + 1}}$ .

$0 (\frac{1}{x^{3} - 7} \cdot \frac{\sqrt{x^{2} + 4} \sqrt{x^{2} + 1}}{\sqrt{x^{2} + 1} \sqrt{x^{2} + 1}})$

Raise $\sqrt{x^{2} + 1}$ to the power of $1$ .

$0 (\frac{1}{x^{3} - 7} \cdot \frac{\sqrt{x^{2} + 4} \sqrt{x^{2} + 1}}{{\sqrt{x^{2} + 1}}^{1} \sqrt{x^{2} + 1}})$

Raise $\sqrt{x^{2} + 1}$ to the power of $1$ .

$0 (\frac{1}{x^{3} - 7} \cdot \frac{\sqrt{x^{2} + 4} \sqrt{x^{2} + 1}}{{\sqrt{x^{2} + 1}}^{1} {\sqrt{x^{2} + 1}}^{1}})$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$0 (\frac{1}{x^{3} - 7} \cdot \frac{\sqrt{x^{2} + 4} \sqrt{x^{2} + 1}}{{\sqrt{x^{2} + 1}}^{1 + 1}})$

Add $1$ and $1$ .

$0 (\frac{1}{x^{3} - 7} \cdot \frac{\sqrt{x^{2} + 4} \sqrt{x^{2} + 1}}{{\sqrt{x^{2} + 1}}^{2}})$

Rewrite ${\sqrt{x^{2} + 1}}^{2}$ as $x^{2} + 1$ .

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

$0 (\frac{1}{x^{3} - 7} \cdot \frac{\sqrt{x^{2} + 4} \sqrt{x^{2} + 1}}{{({(x^{2} + 1)}^{\frac{1}{2}})}^{2}})$

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

$0 (\frac{1}{x^{3} - 7} \cdot \frac{\sqrt{x^{2} + 4} \sqrt{x^{2} + 1}}{{(x^{2} + 1)}^{\frac{1}{2} \cdot 2}})$

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

$0 (\frac{1}{x^{3} - 7} \cdot \frac{\sqrt{x^{2} + 4} \sqrt{x^{2} + 1}}{{(x^{2} + 1)}^{\frac{2}{2}}})$

Cancel the common factor of $2$ .

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Cancel the common factor.

$0 (\frac{1}{x^{3} - 7} \cdot \frac{\sqrt{x^{2} + 4} \sqrt{x^{2} + 1}}{{(x^{2} + 1)}^{\frac{2}{2}}})$

Rewrite the expression.

$0 (\frac{1}{x^{3} - 7} \cdot \frac{\sqrt{x^{2} + 4} \sqrt{x^{2} + 1}}{{(x^{2} + 1)}^{1}})$

Simplify.

$0 (\frac{1}{x^{3} - 7} \cdot \frac{\sqrt{x^{2} + 4} \sqrt{x^{2} + 1}}{x^{2} + 1})$

Step 8

Combine using the product rule for radicals.

$0 (\frac{1}{x^{3} - 7} \cdot \frac{\sqrt{(x^{2} + 4) (x^{2} + 1)}}{x^{2} + 1})$

Step 9

Multiply $\frac{1}{x^{3} - 7}$ by $\frac{\sqrt{(x^{2} + 4) (x^{2} + 1)}}{x^{2} + 1}$ .

$0 \frac{\sqrt{(x^{2} + 4) (x^{2} + 1)}}{(x^{3} - 7) (x^{2} + 1)}$

Step 10

Multiply $0$ by $\frac{\sqrt{(x^{2} + 4) (x^{2} + 1)}}{(x^{3} - 7) (x^{2} + 1)}$ .

$0$