Precalculus Examples

$\log (2 x^{2} + 3 x) = \log (6 x + 2)$

Step 1

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

$2 x^{2} + 3 x > 0$

Step 2

Solve for $x$ .

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Step 2.1

Convert the inequality to an equation.

$2 x^{2} + 3 x = 0$

Step 2.2

Factor $x$ out of $2 x^{2} + 3 x$ .

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Step 2.2.1

Factor $x$ out of $2 x^{2}$ .

$x (2 x) + 3 x = 0$

Step 2.2.2

Factor $x$ out of $3 x$ .

$x (2 x) + x \cdot 3 = 0$

Step 2.2.3

Factor $x$ out of $x (2 x) + x \cdot 3$ .

$x (2 x + 3) = 0$

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 $0$ .

$x = 0$

$2 x + 3 = 0$

Step 2.4

Set $x$ equal to $0$ .

$x = 0$

Step 2.5

Set $2 x + 3$ equal to $0$ and solve for $x$ .

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Step 2.5.1

Set $2 x + 3$ equal to $0$ .

$2 x + 3 = 0$

Step 2.5.2

Solve $2 x + 3 = 0$ for $x$ .

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Step 2.5.2.1

Subtract $3$ from both sides of the equation.

$2 x = - 3$

Step 2.5.2.2

Divide each term in $2 x = - 3$ by $2$ and simplify.

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Step 2.5.2.2.1

Divide each term in $2 x = - 3$ by $2$ .

$\frac{2 x}{2} = \frac{- 3}{2}$

Step 2.5.2.2.2

Simplify the left side.

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Step 2.5.2.2.2.1

Cancel the common factor of $2$ .

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Step 2.5.2.2.2.1.1

Cancel the common factor.

$\frac{2 x}{2} = \frac{- 3}{2}$

Step 2.5.2.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{- 3}{2}$

Step 2.5.2.2.3

Simplify the right side.

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Step 2.5.2.2.3.1

Move the negative in front of the fraction.

$x = - \frac{3}{2}$

Step 2.6

The final solution is all the values that make $x (2 x + 3) = 0$ true.

$x = 0, - \frac{3}{2}$

Step 2.7

Use each root to create test intervals.

$x < - \frac{3}{2}$

$- \frac{3}{2} < x < 0$

$x > 0$

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 $x < - \frac{3}{2}$ to see if it makes the inequality true.

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Step 2.8.1.1

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

$x = - 4$

Step 2.8.1.2

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

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

Step 2.8.1.3

The left side $20$ 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 $- \frac{3}{2} < x < 0$ to see if it makes the inequality true.

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Step 2.8.2.1

Choose a value on the interval $- \frac{3}{2} < x < 0$ and see if this value makes the original inequality true.

$x = - 1$

Step 2.8.2.2

Replace $x$ with $- 1$ in the original inequality.

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

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 $x > 0$ to see if it makes the inequality true.

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Step 2.8.3.1

Choose a value on the interval $x > 0$ and see if this value makes the original inequality true.

$x = 2$

Step 2.8.3.2

Replace $x$ with $2$ in the original inequality.

$2 {(2)}^{2} + 3 (2) > 0$

Step 2.8.3.3

The left side $14$ 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.

$x < - \frac{3}{2}$ True

$- \frac{3}{2} < x < 0$ False

$x > 0$ True

$x < - \frac{3}{2}$ True

$- \frac{3}{2} < x < 0$ False

$x > 0$ True

Step 2.9

The solution consists of all of the true intervals.

$x < - \frac{3}{2}$ or $x > 0$

Step 3

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

Interval Notation:

$(- \infty, - \frac{3}{2}) \cup (0, \infty)$

Set-Builder Notation:

${x | x < - \frac{3}{2}, x > 0}$

Step 4