Precalculus Examples

$\log (x^{2} - 8 x + 15)$

Step 1

Set the argument in $\log (x^{2} - 8 x + 15)$ greater than $0$ to find where the expression is defined.

$x^{2} - 8 x + 15 > 0$

Step 2

Solve for $x$ .

Tap for more steps...

Step 2.1

Convert the inequality to an equation.

$x^{2} - 8 x + 15 = 0$

Step 2.2

Factor $x^{2} - 8 x + 15$ using the AC method.

Tap for more steps...

Step 2.2.1

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $15$ and whose sum is $- 8$ .

$- 5, - 3$

Step 2.2.2

Write the factored form using these integers.

$(x - 5) (x - 3) = 0$

Step 2.3

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x - 5 = 0$

$x - 3 = 0$

Step 2.4

Set $x - 5$ equal to $0$ and solve for $x$ .

Tap for more steps...

Step 2.4.1

Set $x - 5$ equal to $0$ .

$x - 5 = 0$

Step 2.4.2

Add $5$ to both sides of the equation.

$x = 5$

Step 2.5

Set $x - 3$ equal to $0$ and solve for $x$ .

Tap for more steps...

Step 2.5.1

Set $x - 3$ equal to $0$ .

$x - 3 = 0$

Step 2.5.2

Add $3$ to both sides of the equation.

$x = 3$

Step 2.6

The final solution is all the values that make $(x - 5) (x - 3) = 0$ true.

$x = 5, 3$

Step 2.7

Use each root to create test intervals.

$x < 3$

$3 < x < 5$

$x > 5$

Step 2.8

Choose a test value from each interval and plug this value into the original inequality to determine which intervals satisfy the inequality.

Tap for more steps...

Step 2.8.1

Test a value on the interval $x < 3$ to see if it makes the inequality true.

Tap for more steps...

Step 2.8.1.1

Choose a value on the interval $x < 3$ and see if this value makes the original inequality true.

$x = 0$

Step 2.8.1.2

Replace $x$ with $0$ in the original inequality.

${(0)}^{2} - 8 \cdot 0 + 15 > 0$

Step 2.8.1.3

The left side $15$ is greater than the right side $0$ , which means that the given statement is always true.

True

Step 2.8.2

Test a value on the interval $3 < x < 5$ to see if it makes the inequality true.

Tap for more steps...

Step 2.8.2.1

Choose a value on the interval $3 < x < 5$ and see if this value makes the original inequality true.

$x = 4$

Step 2.8.2.2

Replace $x$ with $4$ in the original inequality.

${(4)}^{2} - 8 \cdot 4 + 15 > 0$

Step 2.8.2.3

The left side $- 1$ is not greater than the right side $0$ , which means that the given statement is false.

False

Step 2.8.3

Test a value on the interval $x > 5$ to see if it makes the inequality true.

Tap for more steps...

Step 2.8.3.1

Choose a value on the interval $x > 5$ and see if this value makes the original inequality true.

$x = 8$

Step 2.8.3.2

Replace $x$ with $8$ in the original inequality.

${(8)}^{2} - 8 \cdot 8 + 15 > 0$

Step 2.8.3.3

The left side $15$ is greater than the right side $0$ , which means that the given statement is always true.

True

Step 2.8.4

Compare the intervals to determine which ones satisfy the original inequality.

$x < 3$ True

$3 < x < 5$ False

$x > 5$ True

$x < 3$ True

$3 < x < 5$ False

$x > 5$ True

Step 2.9

The solution consists of all of the true intervals.

$x < 3$ or $x > 5$

Step 3

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

Interval Notation:

$(- \infty, 3) \cup (5, \infty)$

Set-Builder Notation:

${x | x < 3, x > 5}$

Step 4