Solve for x log base 3 of 2x-1> log base 3 of x+2

$\log_{3} (2 x - 1) > \log_{3} (x + 2)$

Step 1

Convert the inequality to an equality.

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

Step 2

Solve the equation.

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

$2 x - 1 = x + 2$

Solve for $x$ .

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

Subtract $x$ from both sides of the equation.

$2 x - 1 - x = 2$

Subtract $x$ from $2 x$ .

$x - 1 = 2$

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

Add $1$ to both sides of the equation.

$x = 2 + 1$

Add $2$ and $1$ .

$x = 3$

Step 3

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

Set the argument in $\log_{3} (\frac{2 x - 1}{x + 2})$ greater than $0$ to find where the expression is defined.

$\frac{2 x - 1}{x + 2} > 0$

Solve for $x$ .

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

$2 x - 1 = 0$

$x + 2 = 0$

Add $1$ to both sides of the equation.

$2 x = 1$

Divide each term in $2 x = 1$ by $2$ and simplify.

Divide each term in $2 x = 1$ by $2$ .

$\frac{2 x}{2} = \frac{1}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x}{2} = \frac{1}{2}$

Divide $x$ by $1$ .

$x = \frac{1}{2}$

Subtract $2$ from both sides of the equation.

$x = - 2$

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

$x = \frac{1}{2}$

$x = - 2$

Consolidate the solutions.

$x = \frac{1}{2}, - 2$

Find the domain of $\frac{2 x - 1}{x + 2}$ .

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

$x + 2 = 0$

Subtract $2$ from both sides of the equation.

$x = - 2$

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

$(- \infty, - 2) \cup (- 2, \infty)$

Use each root to create test intervals.

$x < - 2$

$- 2 < x < \frac{1}{2}$

$x > \frac{1}{2}$

Choose a test value from each interval and plug this value into the original inequality to determine which intervals satisfy the inequality.

Test a value on the interval $x < - 2$ to see if it makes the inequality true.

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

$x = - 4$

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

$\frac{2 (- 4) - 1}{(- 4) + 2} > 0$

The left side $4.5$ is greater than the right side $0$ , which means that the given statement is always true.

True

Test a value on the interval $- 2 < x < \frac{1}{2}$ to see if it makes the inequality true.

Choose a value on the interval $- 2 < x < \frac{1}{2}$ and see if this value makes the original inequality true.

$x = 0$

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

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

The left side $- 0.5$ is not greater than the right side $0$ , which means that the given statement is false.

False

Test a value on the interval $x > \frac{1}{2}$ to see if it makes the inequality true.

Choose a value on the interval $x > \frac{1}{2}$ and see if this value makes the original inequality true.

$x = 3$

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

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

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

True

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

$x < - 2$ True

$- 2 < x < \frac{1}{2}$ False

$x > \frac{1}{2}$ True

$x < - 2$ True

$- 2 < x < \frac{1}{2}$ False

$x > \frac{1}{2}$ True

The solution consists of all of the true intervals.

$x < - 2$ or $x > \frac{1}{2}$

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

$x + 2 = 0$

Subtract $2$ from both sides of the equation.

$x = - 2$

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

$(- \infty, - 2) \cup (\frac{1}{2}, \infty)$

Step 4

The solution consists of all of the true intervals.

$x > 3$

Step 5

The result can be shown in multiple forms.

Inequality Form:

$x > 3$

Interval Notation:

$(3, \infty)$

Step 6

Solve for x log base 3 of 2x-1> log base 3 of x+2

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