Trigonometry Examples

$2 \log_{9} (3 x - 1) = \log_{9} (6 x + 1)$

Step 1

Subtract $\log_{9} (6 x + 1)$ from both sides of the equation.

$2 \log_{9} (3 x - 1) - \log_{9} (6 x + 1) = 0$

Step 2

Simplify $2 \log_{9} (3 x - 1) - \log_{9} (6 x + 1)$ .

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Step 2.1

Simplify $2 \log_{9} (3 x - 1)$ by moving $2$ inside the logarithm.

$\log_{9} ({(3 x - 1)}^{2}) - \log_{9} (6 x + 1) = 0$

Step 2.2

Use the quotient property of logarithms, $\log_{b} (x) - \log_{b} (y) = \log_{b} (\frac{x}{y})$ .

$\log_{9} (\frac{{(3 x - 1)}^{2}}{6 x + 1}) = 0$

Step 3

Set the denominator in $\frac{{(3 x - 1)}^{2}}{6 x + 1}$ equal to $0$ to find where the expression is undefined.

$6 x + 1 = 0$

Step 4

Solve for $x$ .

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Step 4.1

Subtract $1$ from both sides of the equation.

$6 x = - 1$

Step 4.2

Divide each term in $6 x = - 1$ by $6$ and simplify.

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Step 4.2.1

Divide each term in $6 x = - 1$ by $6$ .

$\frac{6 x}{6} = \frac{- 1}{6}$

Step 4.2.2

Simplify the left side.

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Step 4.2.2.1

Cancel the common factor of $6$ .

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Step 4.2.2.1.1

Cancel the common factor.

$\frac{6 x}{6} = \frac{- 1}{6}$

Step 4.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{- 1}{6}$

Step 4.2.3

Simplify the right side.

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Step 4.2.3.1

Move the negative in front of the fraction.

$x = - \frac{1}{6}$

Step 5

Set the argument in $\log_{9} (\frac{{(3 x - 1)}^{2}}{6 x + 1})$ less than or equal to $0$ to find where the expression is undefined.

$\frac{{(3 x - 1)}^{2}}{6 x + 1} \leq 0$

Step 6

Solve for $x$ .

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Step 6.1

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

${(3 x - 1)}^{2} = 0$

$6 x + 1 = 0$

Step 6.2

Set the $3 x - 1$ equal to $0$ .

$3 x - 1 = 0$

Step 6.3

Solve for $x$ .

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Step 6.3.1

Add $1$ to both sides of the equation.

$3 x = 1$

Step 6.3.2

Divide each term in $3 x = 1$ by $3$ and simplify.

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Step 6.3.2.1

Divide each term in $3 x = 1$ by $3$ .

$\frac{3 x}{3} = \frac{1}{3}$

Step 6.3.2.2

Simplify the left side.

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Step 6.3.2.2.1

Cancel the common factor of $3$ .

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Step 6.3.2.2.1.1

Cancel the common factor.

$\frac{3 x}{3} = \frac{1}{3}$

Step 6.3.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{1}{3}$

Step 6.4

Subtract $1$ from both sides of the equation.

$6 x = - 1$

Step 6.5

Divide each term in $6 x = - 1$ by $6$ and simplify.

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Step 6.5.1

Divide each term in $6 x = - 1$ by $6$ .

$\frac{6 x}{6} = \frac{- 1}{6}$

Step 6.5.2

Simplify the left side.

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Step 6.5.2.1

Cancel the common factor of $6$ .

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Step 6.5.2.1.1

Cancel the common factor.

$\frac{6 x}{6} = \frac{- 1}{6}$

Step 6.5.2.1.2

Divide $x$ by $1$ .

$x = \frac{- 1}{6}$

Step 6.5.3

Simplify the right side.

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Step 6.5.3.1

Move the negative in front of the fraction.

$x = - \frac{1}{6}$

Step 6.6

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

$x = \frac{1}{3}$

$x = - \frac{1}{6}$

Step 6.7

Consolidate the solutions.

$x = \frac{1}{3}, - \frac{1}{6}$

Step 6.8

Find the domain of $\frac{{(3 x - 1)}^{2}}{6 x + 1}$ .

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Step 6.8.1

Set the denominator in $\frac{{(3 x - 1)}^{2}}{6 x + 1}$ equal to $0$ to find where the expression is undefined.

$6 x + 1 = 0$

Step 6.8.2

Solve for $x$ .

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Step 6.8.2.1

Subtract $1$ from both sides of the equation.

$6 x = - 1$

Step 6.8.2.2

Divide each term in $6 x = - 1$ by $6$ and simplify.

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Step 6.8.2.2.1

Divide each term in $6 x = - 1$ by $6$ .

$\frac{6 x}{6} = \frac{- 1}{6}$

Step 6.8.2.2.2

Simplify the left side.

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Step 6.8.2.2.2.1

Cancel the common factor of $6$ .

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Step 6.8.2.2.2.1.1

Cancel the common factor.

$\frac{6 x}{6} = \frac{- 1}{6}$

Step 6.8.2.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{- 1}{6}$

Step 6.8.2.2.3

Simplify the right side.

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Step 6.8.2.2.3.1

Move the negative in front of the fraction.

$x = - \frac{1}{6}$

Step 6.8.3

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

$(- \infty, - \frac{1}{6}) \cup (- \frac{1}{6}, \infty)$

Step 6.9

Use each root to create test intervals.

$x < - \frac{1}{6}$

$- \frac{1}{6} < x < \frac{1}{3}$

$x > \frac{1}{3}$

Step 6.10

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 6.10.1

Test a value on the interval $x < - \frac{1}{6}$ to see if it makes the inequality true.

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Step 6.10.1.1

Choose a value on the interval $x < - \frac{1}{6}$ and see if this value makes the original inequality true.

$x = - 3$

Step 6.10.1.2

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

$\frac{{(3 (- 3) - 1)}^{2}}{6 (- 3) + 1} \leq 0$

Step 6.10.1.3

The left side $- 5.88235294$ is less than the right side $0$ , which means that the given statement is always true.

True

Step 6.10.2

Test a value on the interval $- \frac{1}{6} < x < \frac{1}{3}$ to see if it makes the inequality true.

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Step 6.10.2.1

Choose a value on the interval $- \frac{1}{6} < x < \frac{1}{3}$ and see if this value makes the original inequality true.

$x = 0$

Step 6.10.2.2

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

$\frac{{(3 (0) - 1)}^{2}}{6 (0) + 1} \leq 0$

Step 6.10.2.3

The left side $1$ is greater than the right side $0$ , which means that the given statement is false.

False

Step 6.10.3

Test a value on the interval $x > \frac{1}{3}$ to see if it makes the inequality true.

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Step 6.10.3.1

Choose a value on the interval $x > \frac{1}{3}$ and see if this value makes the original inequality true.

$x = 3$

Step 6.10.3.2

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

$\frac{{(3 (3) - 1)}^{2}}{6 (3) + 1} \leq 0$

Step 6.10.3.3

The left side $3.36842105$ is greater than the right side $0$ , which means that the given statement is false.

False

Step 6.10.4

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

$x < - \frac{1}{6}$ True

$- \frac{1}{6} < x < \frac{1}{3}$ False

$x > \frac{1}{3}$ False

$x < - \frac{1}{6}$ True

$- \frac{1}{6} < x < \frac{1}{3}$ False

$x > \frac{1}{3}$ False

Step 6.11

The solution consists of all of the true intervals.

$x < - \frac{1}{6}$ or $x = \frac{1}{3}$

Step 7

The equation is undefined where the denominator equals $0$ , the argument of a square root is less than $0$ , or the argument of a logarithm is less than or equal to $0$ .

$x = - \frac{1}{6}, x = \frac{1}{3}$

Step 8