Precalculus Examples

$f (x) = \log_{7} (x^{2} - 9)$

Step 1

Set the argument in $\log_{7} (x^{2} - 9)$ greater than $0$ to find where the expression is defined.

$x^{2} - 9 > 0$

Step 2

Solve for $x$ .

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Step 2.1

Add $9$ to both sides of the inequality.

$x^{2} > 9$

Step 2.2

Take the specified root of both sides of the inequality to eliminate the exponent on the left side.

$\sqrt{x^{2}} > \sqrt{9}$

Step 2.3

Simplify the equation.

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Step 2.3.1

Simplify the left side.

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Step 2.3.1.1

Pull terms out from under the radical.

$| x | > \sqrt{9}$

Step 2.3.2

Simplify the right side.

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Step 2.3.2.1

Simplify $\sqrt{9}$ .

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Step 2.3.2.1.1

Rewrite $9$ as $3^{2}$ .

$| x | > \sqrt{3^{2}}$

Step 2.3.2.1.2

Pull terms out from under the radical.

$| x | > | 3 |$

Step 2.3.2.1.3

The absolute value is the distance between a number and zero. The distance between $0$ and $3$ is $3$ .

$| x | > 3$

Step 2.4

Write $| x | > 3$ as a piecewise.

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Step 2.4.1

To find the interval for the first piece, find where the inside of the absolute value is non-negative.

$x \geq 0$

Step 2.4.2

In the piece where $x$ is non-negative, remove the absolute value.

$x > 3$

Step 2.4.3

To find the interval for the second piece, find where the inside of the absolute value is negative.

$x < 0$

Step 2.4.4

In the piece where $x$ is negative, remove the absolute value and multiply by $- 1$ .

$- x > 3$

Step 2.4.5

Write as a piecewise.

${\begin{cases} x > 3 & x \geq 0 \\ - x > 3 & x < 0 \end{cases}$

Step 2.5

Find the intersection of $x > 3$ and $x \geq 0$ .

$x > 3$

Step 2.6

Divide each term in $- x > 3$ by $- 1$ and simplify.

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Step 2.6.1

Divide each term in $- x > 3$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- x}{- 1} < \frac{3}{- 1}$

Step 2.6.2

Simplify the left side.

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Step 2.6.2.1

Dividing two negative values results in a positive value.

$\frac{x}{1} < \frac{3}{- 1}$

Step 2.6.2.2

Divide $x$ by $1$ .

$x < \frac{3}{- 1}$

Step 2.6.3

Simplify the right side.

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Step 2.6.3.1

Divide $3$ by $- 1$ .

$x < - 3$

Step 2.7

Find the union of the solutions.

$x < - 3$ or $x > 3$

Step 3

The domain is all values of $x$ that make the expression defined.

Interval Notation:

$(- \infty, - 3) \cup (3, \infty)$

Set-Builder Notation:

${x | x < - 3, x > 3}$

Step 4