Precalculus Examples

$\ln ({(x - 5)}^{2})$

Step 1

Set the argument in $\ln ({(x - 5)}^{2})$ greater than $0$ to find where the expression is defined.

${(x - 5)}^{2} > 0$

Step 2

Solve for $x$ .

Tap for more steps...

Step 2.1

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

$\sqrt{{(x - 5)}^{2}} > \sqrt{0}$

Step 2.2

Simplify the equation.

Tap for more steps...

Step 2.2.1

Simplify the left side.

Tap for more steps...

Step 2.2.1.1

Pull terms out from under the radical.

$| x - 5 | > \sqrt{0}$

Step 2.2.2

Simplify the right side.

Tap for more steps...

Step 2.2.2.1

Simplify $\sqrt{0}$ .

Tap for more steps...

Step 2.2.2.1.1

Rewrite $0$ as $0^{2}$ .

$| x - 5 | > \sqrt{0^{2}}$

Step 2.2.2.1.2

Pull terms out from under the radical.

$| x - 5 | > | 0 |$

Step 2.2.2.1.3

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

$| x - 5 | > 0$

Step 2.3

Write $| x - 5 | > 0$ as a piecewise.

Tap for more steps...

Step 2.3.1

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

$x - 5 \geq 0$

Step 2.3.2

Add $5$ to both sides of the inequality.

$x \geq 5$

Step 2.3.3

In the piece where $x - 5$ is non-negative, remove the absolute value.

$x - 5 > 0$

Step 2.3.4

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

$x - 5 < 0$

Step 2.3.5

Add $5$ to both sides of the inequality.

$x < 5$

Step 2.3.6

In the piece where $x - 5$ is negative, remove the absolute value and multiply by $- 1$ .

$- (x - 5) > 0$

Step 2.3.7

Write as a piecewise.

${\begin{cases} x - 5 > 0 & x \geq 5 \\ - (x - 5) > 0 & x < 5 \end{cases}$

Step 2.3.8

Simplify $- (x - 5) > 0$ .

Tap for more steps...

Step 2.3.8.1

Apply the distributive property.

${\begin{cases} x - 5 > 0 & x \geq 5 \\ - x - - 5 > 0 & x < 5 \end{cases}$

Step 2.3.8.2

Multiply $- 1$ by $- 5$ .

${\begin{cases} x - 5 > 0 & x \geq 5 \\ - x + 5 > 0 & x < 5 \end{cases}$

Step 2.4

Add $5$ to both sides of the inequality.

$x > 5$

Step 2.5

Solve $- x + 5 > 0$ for $x$ .

Tap for more steps...

Step 2.5.1

Subtract $5$ from both sides of the inequality.

$- x > - 5$

Step 2.5.2

Divide each term in $- x > - 5$ by $- 1$ and simplify.

Tap for more steps...

Step 2.5.2.1

Divide each term in $- x > - 5$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- x}{- 1} < \frac{- 5}{- 1}$

Step 2.5.2.2

Simplify the left side.

Tap for more steps...

Step 2.5.2.2.1

Dividing two negative values results in a positive value.

$\frac{x}{1} < \frac{- 5}{- 1}$

Step 2.5.2.2.2

Divide $x$ by $1$ .

$x < \frac{- 5}{- 1}$

Step 2.5.2.3

Simplify the right side.

Tap for more steps...

Step 2.5.2.3.1

Divide $- 5$ by $- 1$ .

$x < 5$

Step 2.6

Find the union of the solutions.

$x < 5$ or $x > 5$

Step 3

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

Interval Notation:

$(- \infty, 5) \cup (5, \infty)$

Set-Builder Notation:

${x | x \neq 5}$

Step 4