Find the Domain log of (x+3)/(x-2)

$\log (\frac{x + 3}{x - 2})$

Step 1

Set the argument in $\log (\frac{x + 3}{x - 2})$ greater than $0$ to find where the expression is defined.

$\frac{x + 3}{x - 2} > 0$

Step 2

Solve for $x$ .

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

$x + 3 = 0$

$x - 2 = 0$

Subtract $3$ from both sides of the equation.

$x = - 3$

Add $2$ to both sides of the equation.

$x = 2$

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

$x = - 3$

$x = 2$

Consolidate the solutions.

$x = - 3, 2$

Find the domain of $\frac{x + 3}{x - 2}$ .

Set the denominator in $\frac{x + 3}{x - 2}$ equal to $0$ to find where the expression is undefined.

$x - 2 = 0$

Add $2$ to both sides of the equation.

$x = 2$

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

$(- \infty, 2) \cup (2, \infty)$

Use each root to create test intervals.

$x < - 3$

$- 3 < x < 2$

$x > 2$

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

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

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

$x = - 6$

Replace $x$ with $- 6$ in the original inequality.

$\frac{(- 6) + 3}{(- 6) - 2} > 0$

The left side $0.375$ is greater than the right side $0$ , which means that the given statement is always true.

True

Test a value on the interval $- 3 < x < 2$ to see if it makes the inequality true.

Choose a value on the interval $- 3 < x < 2$ and see if this value makes the original inequality true.

$x = 0$

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

$\frac{(0) + 3}{(0) - 2} > 0$

The left side $- 1.5$ is not greater than the right side $0$ , which means that the given statement is false.

False

Test a value on the interval $x > 2$ to see if it makes the inequality true.

Choose a value on the interval $x > 2$ and see if this value makes the original inequality true.

$x = 4$

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

$\frac{(4) + 3}{(4) - 2} > 0$

The left side $3.5$ is greater than the right side $0$ , which means that the given statement is always true.

True

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

$x < - 3$ True

$- 3 < x < 2$ False

$x > 2$ True

$x < - 3$ True

$- 3 < x < 2$ False

$x > 2$ True

The solution consists of all of the true intervals.

$x < - 3$ or $x > 2$

Step 3

Set the denominator in $\frac{x + 3}{x - 2}$ equal to $0$ to find where the expression is undefined.

$x - 2 = 0$

Step 4

Add $2$ to both sides of the equation.

$x = 2$

Step 5

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

Interval Notation:

$(- \infty, - 3) \cup (2, \infty)$

Set-Builder Notation:

${x | x < - 3, x > 2}$

Step 6

Find the Domain log of (x+3)/(x-2)

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