Precalculus Examples

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

Step 1

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

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

Step 2

Solve for $x$ .

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Step 2.1

To remove the radical on the left side of the inequality, square both sides of the inequality.

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

Step 2.2

Simplify each side of the inequality.

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Step 2.2.1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{\frac{x + 3}{x - 2}}$ as ${(\frac{x + 3}{x - 2})}^{\frac{1}{2}}$ .

${({(\frac{x + 3}{x - 2})}^{\frac{1}{2}})}^{2} > 0^{2}$

Step 2.2.2

Simplify the left side.

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Step 2.2.2.1

Simplify ${({(\frac{x + 3}{x - 2})}^{\frac{1}{2}})}^{2}$ .

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Step 2.2.2.1.1

Multiply the exponents in ${({(\frac{x + 3}{x - 2})}^{\frac{1}{2}})}^{2}$ .

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Step 2.2.2.1.1.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

${(\frac{x + 3}{x - 2})}^{\frac{1}{2} \cdot 2} > 0^{2}$

Step 2.2.2.1.1.2

Cancel the common factor of $2$ .

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Step 2.2.2.1.1.2.1

Cancel the common factor.

${(\frac{x + 3}{x - 2})}^{\frac{1}{2} \cdot 2} > 0^{2}$

Step 2.2.2.1.1.2.2

Rewrite the expression.

${(\frac{x + 3}{x - 2})}^{1} > 0^{2}$

Step 2.2.2.1.2

Simplify.

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

Step 2.2.3

Simplify the right side.

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Step 2.2.3.1

Raising $0$ to any positive power yields $0$ .

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

Step 2.3

Solve for $x$ .

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Step 2.3.1

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

$x + 3 = 0$

$x - 2 = 0$

Step 2.3.2

Subtract $3$ from both sides of the equation.

$x = - 3$

Step 2.3.3

Add $2$ to both sides of the equation.

$x = 2$

Step 2.3.4

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

$x = - 3$

$x = 2$

Step 2.3.5

Consolidate the solutions.

$x = - 3, 2$

Step 2.4

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

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Step 2.4.1

Set the radicand in $\sqrt{\frac{x + 3}{x - 2}}$ greater than or equal to $0$ to find where the expression is defined.

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

Step 2.4.2

Solve for $x$ .

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Step 2.4.2.1

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

$x + 3 = 0$

$x - 2 = 0$

Step 2.4.2.2

Subtract $3$ from both sides of the equation.

$x = - 3$

Step 2.4.2.3

Add $2$ to both sides of the equation.

$x = 2$

Step 2.4.2.4

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

$x = - 3$

$x = 2$

Step 2.4.2.5

Consolidate the solutions.

$x = - 3, 2$

Step 2.4.2.6

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

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Step 2.4.2.6.1

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

$x - 2 = 0$

Step 2.4.2.6.2

Add $2$ to both sides of the equation.

$x = 2$

Step 2.4.2.6.3

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

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

Step 2.4.2.7

Use each root to create test intervals.

$x < - 3$

$- 3 < x < 2$

$x > 2$

Step 2.4.2.8

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 2.4.2.8.1

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

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Step 2.4.2.8.1.1

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

$x = - 6$

Step 2.4.2.8.1.2

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

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

Step 2.4.2.8.1.3

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

True

Step 2.4.2.8.2

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

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Step 2.4.2.8.2.1

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

$x = 0$

Step 2.4.2.8.2.2

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

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

Step 2.4.2.8.2.3

The left side $- 1.5$ is less than the right side $0$ , which means that the given statement is false.

False

Step 2.4.2.8.3

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

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Step 2.4.2.8.3.1

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

$x = 4$

Step 2.4.2.8.3.2

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

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

Step 2.4.2.8.3.3

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

True

Step 2.4.2.8.4

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

Step 2.4.2.9

The solution consists of all of the true intervals.

$x \leq - 3$ or $x > 2$

Step 2.4.3

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

$x - 2 = 0$

Step 2.4.4

Add $2$ to both sides of the equation.

$x = 2$

Step 2.4.5

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

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

Step 2.5

Use each root to create test intervals.

$x < - 3$

$- 3 < x < 2$

$x > 2$

Step 2.6

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 2.6.1

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

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Step 2.6.1.1

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

$x = - 6$

Step 2.6.1.2

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

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

Step 2.6.1.3

The left side $0.61237243$ is greater than the right side $0$ , which means that the given statement is always true.

True

Step 2.6.2

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

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Step 2.6.2.1

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

$x = 0$

Step 2.6.2.2

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

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

Step 2.6.2.3

The left side is not equal to the right side, which means that the given statement is false.

False

Step 2.6.3

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

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Step 2.6.3.1

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

$x = 4$

Step 2.6.3.2

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

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

Step 2.6.3.3

The left side $1.87082869$ is greater than the right side $0$ , which means that the given statement is always true.

True

Step 2.6.4

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

Step 2.7

The solution consists of all of the true intervals.

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

Step 3

Set the radicand in $\sqrt{\frac{x + 3}{x - 2}}$ greater than or equal to $0$ to find where the expression is defined.

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

Step 4

Solve for $x$ .

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Step 4.1

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

$x + 3 = 0$

$x - 2 = 0$

Step 4.2

Subtract $3$ from both sides of the equation.

$x = - 3$

Step 4.3

Add $2$ to both sides of the equation.

$x = 2$

Step 4.4

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

$x = - 3$

$x = 2$

Step 4.5

Consolidate the solutions.

$x = - 3, 2$

Step 4.6

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

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Step 4.6.1

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

$x - 2 = 0$

Step 4.6.2

Add $2$ to both sides of the equation.

$x = 2$

Step 4.6.3

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

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

Step 4.7

Use each root to create test intervals.

$x < - 3$

$- 3 < x < 2$

$x > 2$

Step 4.8

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 4.8.1

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

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Step 4.8.1.1

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

$x = - 6$

Step 4.8.1.2

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

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

Step 4.8.1.3

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

True

Step 4.8.2

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

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Step 4.8.2.1

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

$x = 0$

Step 4.8.2.2

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

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

Step 4.8.2.3

The left side $- 1.5$ is less than the right side $0$ , which means that the given statement is false.

False

Step 4.8.3

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

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Step 4.8.3.1

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

$x = 4$

Step 4.8.3.2

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

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

Step 4.8.3.3

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

True

Step 4.8.4

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

Step 4.9

The solution consists of all of the true intervals.

$x \leq - 3$ or $x > 2$

Step 5

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

$x - 2 = 0$

Step 6

Add $2$ to both sides of the equation.

$x = 2$

Step 7

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 8