Trigonometry Examples

$\log_{8} (x^{2} - x) > \log_{8} (42)$

Step 1

Convert the inequality to an equality.

$\log_{8} (x^{2} - x) = \log_{8} (42)$

Step 2

Solve the equation.

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Step 2.1

For the equation to be equal, the argument of the logarithms on both sides of the equation must be equal.

$x^{2} - x = 42$

Step 2.2

Solve for $x$ .

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Step 2.2.1

Subtract $42$ from both sides of the equation.

$x^{2} - x - 42 = 0$

Step 2.2.2

Factor $x^{2} - x - 42$ using the AC method.

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Step 2.2.2.1

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 42$ and whose sum is $- 1$ .

$- 7, 6$

Step 2.2.2.2

Write the factored form using these integers.

$(x - 7) (x + 6) = 0$

Step 2.2.3

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

$x - 7 = 0$

$x + 6 = 0$

Step 2.2.4

Set $x - 7$ equal to $0$ and solve for $x$ .

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Step 2.2.4.1

Set $x - 7$ equal to $0$ .

$x - 7 = 0$

Step 2.2.4.2

Add $7$ to both sides of the equation.

$x = 7$

Step 2.2.5

Set $x + 6$ equal to $0$ and solve for $x$ .

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Step 2.2.5.1

Set $x + 6$ equal to $0$ .

$x + 6 = 0$

Step 2.2.5.2

Subtract $6$ from both sides of the equation.

$x = - 6$

Step 2.2.6

The final solution is all the values that make $(x - 7) (x + 6) = 0$ true.

$x = 7, - 6$

Step 3

Find the domain of $\log_{8} (\frac{x (x - 1)}{42})$ .

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Step 3.1

Set the argument in $\log_{8} (\frac{x (x - 1)}{42})$ greater than $0$ to find where the expression is defined.

$\frac{x (x - 1)}{42} > 0$

Step 3.2

Solve for $x$ .

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Step 3.2.1

Multiply both sides by $42$ .

$\frac{x (x - 1)}{42} \cdot 42 > 0 \cdot 42$

Step 3.2.2

Simplify.

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Step 3.2.2.1

Simplify the left side.

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Step 3.2.2.1.1

Simplify $\frac{x (x - 1)}{42} \cdot 42$ .

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Step 3.2.2.1.1.1

Simplify terms.

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Step 3.2.2.1.1.1.1

Cancel the common factor of $42$ .

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Step 3.2.2.1.1.1.1.1

Cancel the common factor.

$\frac{x (x - 1)}{42} \cdot 42 > 0 \cdot 42$

Step 3.2.2.1.1.1.1.2

Rewrite the expression.

$x (x - 1) > 0 \cdot 42$

Step 3.2.2.1.1.1.2

Apply the distributive property.

$x \cdot x + x \cdot - 1 > 0 \cdot 42$

Step 3.2.2.1.1.1.3

Simplify the expression.

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Step 3.2.2.1.1.1.3.1

Multiply $x$ by $x$ .

$x^{2} + x \cdot - 1 > 0 \cdot 42$

Step 3.2.2.1.1.1.3.2

Move $- 1$ to the left of $x$ .

$x^{2} - 1 \cdot x > 0 \cdot 42$

Step 3.2.2.1.1.2

Rewrite $- 1 x$ as $- x$ .

$x^{2} - x > 0 \cdot 42$

Step 3.2.2.2

Simplify the right side.

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Step 3.2.2.2.1

Multiply $0$ by $42$ .

$x^{2} - x > 0$

Step 3.2.3

Solve for $x$ .

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Step 3.2.3.1

Convert the inequality to an equation.

$x^{2} - x = 0$

Step 3.2.3.2

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

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Step 3.2.3.2.1

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

$x \cdot x - x = 0$

Step 3.2.3.2.2

Factor $x$ out of $- x$ .

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

Step 3.2.3.2.3

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

$x (x - 1) = 0$

Step 3.2.3.3

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 3.2.3.4

Set $x$ equal to $0$ .

$x = 0$

Step 3.2.3.5

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

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Step 3.2.3.5.1

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

$x - 1 = 0$

Step 3.2.3.5.2

Add $1$ to both sides of the equation.

$x = 1$

Step 3.2.3.6

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

$x = 0, 1$

Step 3.2.4

Use each root to create test intervals.

$x < 0$

$0 < x < 1$

$x > 1$

Step 3.2.5

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 3.2.5.1

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

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Step 3.2.5.1.1

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

$x = - 2$

Step 3.2.5.1.2

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

$\frac{(- 2) ((- 2) - 1)}{42} > 0$

Step 3.2.5.1.3

The left side $0.\overline{142857}$ is greater than the right side $0$ , which means that the given statement is always true.

True

Step 3.2.5.2

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

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Step 3.2.5.2.1

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

$x = 0.5$

Step 3.2.5.2.2

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

$\frac{(0.5) ((0.5) - 1)}{42} > 0$

Step 3.2.5.2.3

The left side $- 0.00595238$ is not greater than the right side $0$ , which means that the given statement is false.

False

Step 3.2.5.3

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

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Step 3.2.5.3.1

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

$x = 4$

Step 3.2.5.3.2

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

$\frac{(4) ((4) - 1)}{42} > 0$

Step 3.2.5.3.3

The left side $0.\overline{285714}$ is greater than the right side $0$ , which means that the given statement is always true.

True

Step 3.2.5.4

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 3.2.6

The solution consists of all of the true intervals.

$x < 0$ or $x > 1$

Step 3.3

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

$(- \infty, 0) \cup (1, \infty)$

Step 4

Use each root to create test intervals.

$x < - 6$

$- 6 < x < 0$

$0 < x < 1$

$1 < x < 7$

$x > 7$

Step 5

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 5.1

Test a value on the interval $x < - 6$ to see if it makes the inequality true.

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Step 5.1.1

Choose a value on the interval $x < - 6$ and see if this value makes the original inequality true.

$x = - 8$

Step 5.1.2

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

$\log_{8} ({(- 8)}^{2} - (- 8)) > \log_{8} (42)$

Step 5.1.3

The left side $2.05664166$ is greater than the right side $1.79743914$ , which means that the given statement is always true.

True

Step 5.2

Test a value on the interval $- 6 < x < 0$ to see if it makes the inequality true.

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Step 5.2.1

Choose a value on the interval $- 6 < x < 0$ and see if this value makes the original inequality true.

$x = - 3$

Step 5.2.2

Replace $x$ with $- 3$ in the original inequality.

$\log_{8} ({(- 3)}^{2} - (- 3)) > \log_{8} (42)$

Step 5.2.3

The left side $1.1949875$ is not greater than the right side $1.79743914$ , which means that the given statement is false.

False

Step 5.3

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

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Step 5.3.1

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

$x = 0.5$

Step 5.3.2

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

$\log_{8} ({(0.5)}^{2} - (0.5)) > \log_{8} (42)$

Step 5.3.3

Determine if the inequality is true.

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Step 5.3.3.1

The equation cannot be solved because it is undefined.

$Undefined$

Step 5.3.3.2

The left side has no solution, which means that the given statement is false.

False

Step 5.4

Test a value on the interval $1 < x < 7$ to see if it makes the inequality true.

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Step 5.4.1

Choose a value on the interval $1 < x < 7$ and see if this value makes the original inequality true.

$x = 4$

Step 5.4.2

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

$\log_{8} ({(4)}^{2} - (4)) > \log_{8} (42)$

Step 5.4.3

The left side $1.1949875$ is not greater than the right side $1.79743914$ , which means that the given statement is false.

False

Step 5.5

Test a value on the interval $x > 7$ to see if it makes the inequality true.

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Step 5.5.1

Choose a value on the interval $x > 7$ and see if this value makes the original inequality true.

$x = 10$

Step 5.5.2

Replace $x$ with $10$ in the original inequality.

$\log_{8} ({(10)}^{2} - (10)) > \log_{8} (42)$

Step 5.5.3

The left side $2.16395103$ is greater than the right side $1.79743914$ , which means that the given statement is always true.

True

Step 5.6

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

$x < - 6$ True

$- 6 < x < 0$ False

$0 < x < 1$ False

$1 < x < 7$ False

$x > 7$ True

$x < - 6$ True

$- 6 < x < 0$ False

$0 < x < 1$ False

$1 < x < 7$ False

$x > 7$ True

Step 6

The solution consists of all of the true intervals.

$x < - 6$ or $x > 7$

Step 7

The result can be shown in multiple forms.

Inequality Form:

$x < - 6 or x > 7$

Interval Notation:

$(- \infty, - 6) \cup (7, \infty)$

Step 8