Algebra Examples

$\log_{5} (3 x + 2) < \log_{5} (2 x + 5)$

Step 1

Convert the inequality to an equality.

$\log_{5} (3 x + 2) = \log_{5} (2 x + 5)$

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.

$3 x + 2 = 2 x + 5$

Step 2.2

Solve for $x$ .

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Step 2.2.1

Move all terms containing $x$ to the left side of the equation.

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Step 2.2.1.1

Subtract $2 x$ from both sides of the equation.

$3 x + 2 - 2 x = 5$

Step 2.2.1.2

Subtract $2 x$ from $3 x$ .

$x + 2 = 5$

Step 2.2.2

Move all terms not containing $x$ to the right side of the equation.

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Step 2.2.2.1

Subtract $2$ from both sides of the equation.

$x = 5 - 2$

Step 2.2.2.2

Subtract $2$ from $5$ .

$x = 3$

Step 3

Find the domain of $\log_{5} (\frac{3 x + 2}{2 x + 5})$ .

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Step 3.1

Set the argument in $\log_{5} (\frac{3 x + 2}{2 x + 5})$ greater than $0$ to find where the expression is defined.

$\frac{3 x + 2}{2 x + 5} > 0$

Step 3.2

Solve for $x$ .

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Step 3.2.1

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

$3 x + 2 = 0$

$2 x + 5 = 0$

Step 3.2.2

Subtract $2$ from both sides of the equation.

$3 x = - 2$

Step 3.2.3

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

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Step 3.2.3.1

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

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

Step 3.2.3.2

Simplify the left side.

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Step 3.2.3.2.1

Cancel the common factor of $3$ .

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Step 3.2.3.2.1.1

Cancel the common factor.

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

Step 3.2.3.2.1.2

Divide $x$ by $1$ .

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

Step 3.2.3.3

Simplify the right side.

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Step 3.2.3.3.1

Move the negative in front of the fraction.

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

Step 3.2.4

Subtract $5$ from both sides of the equation.

$2 x = - 5$

Step 3.2.5

Divide each term in $2 x = - 5$ by $2$ and simplify.

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Step 3.2.5.1

Divide each term in $2 x = - 5$ by $2$ .

$\frac{2 x}{2} = \frac{- 5}{2}$

Step 3.2.5.2

Simplify the left side.

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Step 3.2.5.2.1

Cancel the common factor of $2$ .

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Step 3.2.5.2.1.1

Cancel the common factor.

$\frac{2 x}{2} = \frac{- 5}{2}$

Step 3.2.5.2.1.2

Divide $x$ by $1$ .

$x = \frac{- 5}{2}$

Step 3.2.5.3

Simplify the right side.

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Step 3.2.5.3.1

Move the negative in front of the fraction.

$x = - \frac{5}{2}$

Step 3.2.6

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

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

$x = - \frac{5}{2}$

Step 3.2.7

Consolidate the solutions.

$x = - \frac{2}{3}, - \frac{5}{2}$

Step 3.2.8

Find the domain of $\frac{3 x + 2}{2 x + 5}$ .

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Step 3.2.8.1

Set the denominator in $\frac{3 x + 2}{2 x + 5}$ equal to $0$ to find where the expression is undefined.

$2 x + 5 = 0$

Step 3.2.8.2

Solve for $x$ .

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Step 3.2.8.2.1

Subtract $5$ from both sides of the equation.

$2 x = - 5$

Step 3.2.8.2.2

Divide each term in $2 x = - 5$ by $2$ and simplify.

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Step 3.2.8.2.2.1

Divide each term in $2 x = - 5$ by $2$ .

$\frac{2 x}{2} = \frac{- 5}{2}$

Step 3.2.8.2.2.2

Simplify the left side.

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Step 3.2.8.2.2.2.1

Cancel the common factor of $2$ .

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Step 3.2.8.2.2.2.1.1

Cancel the common factor.

$\frac{2 x}{2} = \frac{- 5}{2}$

Step 3.2.8.2.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{- 5}{2}$

Step 3.2.8.2.2.3

Simplify the right side.

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Step 3.2.8.2.2.3.1

Move the negative in front of the fraction.

$x = - \frac{5}{2}$

Step 3.2.8.3

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

$(- \infty, - \frac{5}{2}) \cup (- \frac{5}{2}, \infty)$

Step 3.2.9

Use each root to create test intervals.

$x < - \frac{5}{2}$

$- \frac{5}{2} < x < - \frac{2}{3}$

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

Step 3.2.10

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 3.2.10.1

Test a value on the interval $x < - \frac{5}{2}$ to see if it makes the inequality true.

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Step 3.2.10.1.1

Choose a value on the interval $x < - \frac{5}{2}$ and see if this value makes the original inequality true.

$x = - 5$

Step 3.2.10.1.2

Replace $x$ with $- 5$ in the original inequality.

$\frac{3 (- 5) + 2}{2 (- 5) + 5} > 0$

Step 3.2.10.1.3

The left side $2.6$ is greater than the right side $0$ , which means that the given statement is always true.

True

Step 3.2.10.2

Test a value on the interval $- \frac{5}{2} < x < - \frac{2}{3}$ to see if it makes the inequality true.

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Step 3.2.10.2.1

Choose a value on the interval $- \frac{5}{2} < x < - \frac{2}{3}$ and see if this value makes the original inequality true.

$x = - 2$

Step 3.2.10.2.2

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

$\frac{3 (- 2) + 2}{2 (- 2) + 5} > 0$

Step 3.2.10.2.3

The left side $- 4$ is not greater than the right side $0$ , which means that the given statement is false.

False

Step 3.2.10.3

Test a value on the interval $x > - \frac{2}{3}$ to see if it makes the inequality true.

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Step 3.2.10.3.1

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

$x = 0$

Step 3.2.10.3.2

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

$\frac{3 (0) + 2}{2 (0) + 5} > 0$

Step 3.2.10.3.3

The left side $0.4$ is greater than the right side $0$ , which means that the given statement is always true.

True

Step 3.2.10.4

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

$x < - \frac{5}{2}$ True

$- \frac{5}{2} < x < - \frac{2}{3}$ False

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

$x < - \frac{5}{2}$ True

$- \frac{5}{2} < x < - \frac{2}{3}$ False

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

Step 3.2.11

The solution consists of all of the true intervals.

$x < - \frac{5}{2}$ or $x > - \frac{2}{3}$

Step 3.3

Set the denominator in $\frac{3 x + 2}{2 x + 5}$ equal to $0$ to find where the expression is undefined.

$2 x + 5 = 0$

Step 3.4

Solve for $x$ .

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Step 3.4.1

Subtract $5$ from both sides of the equation.

$2 x = - 5$

Step 3.4.2

Divide each term in $2 x = - 5$ by $2$ and simplify.

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Step 3.4.2.1

Divide each term in $2 x = - 5$ by $2$ .

$\frac{2 x}{2} = \frac{- 5}{2}$

Step 3.4.2.2

Simplify the left side.

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Step 3.4.2.2.1

Cancel the common factor of $2$ .

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Step 3.4.2.2.1.1

Cancel the common factor.

$\frac{2 x}{2} = \frac{- 5}{2}$

Step 3.4.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{- 5}{2}$

Step 3.4.2.3

Simplify the right side.

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Step 3.4.2.3.1

Move the negative in front of the fraction.

$x = - \frac{5}{2}$

Step 3.5

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

$(- \infty, - \frac{5}{2}) \cup (- \frac{2}{3}, \infty)$

Step 4

Use each root to create test intervals.

$x < - \frac{5}{2}$

$- \frac{5}{2} < x < - \frac{2}{3}$

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

$x > 3$

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 $x < - \frac{5}{2}$ to see if it makes the inequality true.

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Step 5.1.1

Choose a value on the interval $x < - \frac{5}{2}$ and see if this value makes the original inequality true.

$x = - 5$

Step 5.1.2

Replace $x$ with $- 5$ in the original inequality.

$\log_{5} (3 (- 5) + 2) < \log_{5} (2 (- 5) + 5)$

Step 5.1.3

Determine if the inequality is true.

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Step 5.1.3.1

The equation cannot be solved because it is undefined.

$Undefined$

Step 5.1.3.2

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

False

Step 5.2

Test a value on the interval $- \frac{5}{2} < x < - \frac{2}{3}$ to see if it makes the inequality true.

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Step 5.2.1

Choose a value on the interval $- \frac{5}{2} < x < - \frac{2}{3}$ and see if this value makes the original inequality true.

$x = - 2$

Step 5.2.2

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

$\log_{5} (3 (- 2) + 2) < \log_{5} (2 (- 2) + 5)$

Step 5.2.3

Determine if the inequality is true.

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Step 5.2.3.1

The equation cannot be solved because it is undefined.

$Undefined$

Step 5.2.3.2

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

False

Step 5.3

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

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Step 5.3.1

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

$x = 0$

Step 5.3.2

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

$\log_{5} (3 (0) + 2) < \log_{5} (2 (0) + 5)$

Step 5.3.3

The left side $0.43067655$ is less than the right side $1$ , which means that the given statement is always true.

True

Step 5.4

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

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Step 5.4.1

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

$x = 6$

Step 5.4.2

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

$\log_{5} (3 (6) + 2) < \log_{5} (2 (6) + 5)$

Step 5.4.3

The left side $1.86135311$ is not less than the right side $1.76037442$ , which means that the given statement is false.

False

Step 5.5

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

$x < - \frac{5}{2}$ False

$- \frac{5}{2} < x < - \frac{2}{3}$ False

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

$x > 3$ False

$x < - \frac{5}{2}$ False

$- \frac{5}{2} < x < - \frac{2}{3}$ False

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

$x > 3$ False

Step 6

The solution consists of all of the true intervals.

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

Step 7

The result can be shown in multiple forms.

Inequality Form:

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

Interval Notation:

$(- \frac{2}{3}, 3)$

Step 8