Finite Math Examples

$\frac{1}{3} \cdot (2 \log (x - 1) - \log (x - 4) - \log (x^{2} - x) - 3 \log (x + 3))$

Step 1

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

$x - 1 > 0$

Step 2

Add $1$ to both sides of the inequality.

$x > 1$

Step 3

Set the argument in $\log (x - 4)$ greater than $0$ to find where the expression is defined.

$x - 4 > 0$

Step 4

Add $4$ to both sides of the inequality.

$x > 4$

Step 5

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

$x^{2} - x > 0$

Step 6

Solve for $x$ .

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Step 6.1

Convert the inequality to an equation.

$x^{2} - x = 0$

Step 6.2

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

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Step 6.2.1

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

$x \cdot x - x = 0$

Step 6.2.2

Factor $x$ out of $- x$ .

$x \cdot x + x \cdot - 1 = 0$

Step 6.2.3

Factor $x$ out of $x \cdot x + x \cdot - 1$ .

$x (x - 1) = 0$

Step 6.3

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

$x = 0$

$x - 1 = 0$

Step 6.4

Set $x$ equal to $0$ .

$x = 0$

Step 6.5

Set $x - 1$ equal to $0$ and solve for $x$ .

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Step 6.5.1

Set $x - 1$ equal to $0$ .

$x - 1 = 0$

Step 6.5.2

Add $1$ to both sides of the equation.

$x = 1$

Step 6.6

The final solution is all the values that make $x (x - 1) = 0$ true.

$x = 0, 1$

Step 6.7

Use each root to create test intervals.

$x < 0$

$0 < x < 1$

$x > 1$

Step 6.8

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 6.8.1

Test a value on the interval $x < 0$ to see if it makes the inequality true.

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Step 6.8.1.1

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

$x = - 2$

Step 6.8.1.2

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

${(- 2)}^{2} - (- 2) > 0$

Step 6.8.1.3

The left side $6$ is greater than the right side $0$ , which means that the given statement is always true.

True

Step 6.8.2

Test a value on the interval $0 < x < 1$ to see if it makes the inequality true.

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Step 6.8.2.1

Choose a value on the interval $0 < x < 1$ and see if this value makes the original inequality true.

$x = 0.5$

Step 6.8.2.2

Replace $x$ with $0.5$ in the original inequality.

${(0.5)}^{2} - (0.5) > 0$

Step 6.8.2.3

The left side $- 0.25$ is not greater than the right side $0$ , which means that the given statement is false.

False

Step 6.8.3

Test a value on the interval $x > 1$ to see if it makes the inequality true.

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Step 6.8.3.1

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

$x = 4$

Step 6.8.3.2

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

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

Step 6.8.3.3

The left side $12$ is greater than the right side $0$ , which means that the given statement is always true.

True

Step 6.8.4

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

$x < 0$ True

$0 < x < 1$ False

$x > 1$ True

$x < 0$ True

$0 < x < 1$ False

$x > 1$ True

Step 6.9

The solution consists of all of the true intervals.

$x < 0$ or $x > 1$

Step 7

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

$x + 3 > 0$

Step 8

Subtract $3$ from both sides of the inequality.

$x > - 3$

Step 9

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

Interval Notation:

$(4, \infty)$

Set-Builder Notation:

${x | x > 4}$

Step 10