Algebra Examples

$f (x) = \log (\frac{x + 5}{x - 1})$

Step 1

Set the argument in $\log (\frac{x + 5}{x - 1})$ greater than $0$ to find where the expression is defined.

$\frac{x + 5}{x - 1} > 0$

Step 2

Solve for $x$ .

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Step 2.1

Find all the values where the expression switches from negative to positive by setting each factor equal to $0$ and solving.

$x + 5 = 0$

$x - 1 = 0$

Step 2.2

Subtract $5$ from both sides of the equation.

$x = - 5$

Step 2.3

Add $1$ to both sides of the equation.

$x = 1$

Step 2.4

Solve for each factor to find the values where the absolute value expression goes from negative to positive.

$x = - 5$

$x = 1$

Step 2.5

Consolidate the solutions.

$x = - 5, 1$

Step 2.6

Find the domain of $\frac{x + 5}{x - 1}$ .

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Step 2.6.1

Set the denominator in $\frac{x + 5}{x - 1}$ equal to $0$ to find where the expression is undefined.

$x - 1 = 0$

Step 2.6.2

Add $1$ to both sides of the equation.

$x = 1$

Step 2.6.3

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

$(- \infty, 1) \cup (1, \infty)$

Step 2.7

Use each root to create test intervals.

$x < - 5$

$- 5 < x < 1$

$x > 1$

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.

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Step 2.8.1

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

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Step 2.8.1.1

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

$x = - 8$

Step 2.8.1.2

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

$\frac{(- 8) + 5}{(- 8) - 1} > 0$

Step 2.8.1.3

The left side $0.\overline{3}$ 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 $- 5 < x < 1$ to see if it makes the inequality true.

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Step 2.8.2.1

Choose a value on the interval $- 5 < x < 1$ and see if this value makes the original inequality true.

$x = 0$

Step 2.8.2.2

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

$\frac{(0) + 5}{(0) - 1} > 0$

Step 2.8.2.3

The left side $- 5$ 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 $x > 1$ to see if it makes the inequality true.

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Step 2.8.3.1

Choose a value on the interval $x > 1$ and see if this value makes the original inequality true.

$x = 4$

Step 2.8.3.2

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

$\frac{(4) + 5}{(4) - 1} > 0$

Step 2.8.3.3

The left side $3$ 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.

$x < - 5$ True

$- 5 < x < 1$ False

$x > 1$ True

$x < - 5$ True

$- 5 < x < 1$ False

$x > 1$ True

Step 2.9

The solution consists of all of the true intervals.

$x < - 5$ or $x > 1$

Step 3

Set the denominator in $\frac{x + 5}{x - 1}$ equal to $0$ to find where the expression is undefined.

$x - 1 = 0$

Step 4

Add $1$ to both sides of the equation.

$x = 1$

Step 5

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

Interval Notation:

$(- \infty, - 5) \cup (1, \infty)$

Set-Builder Notation:

${x | x < - 5, x > 1}$

Step 6