Solve the Inequality for x log of x> = log of 2x-1- log of 3

$\log (x) \geq \log (2 x - 1) - \log (3)$

Step 1

Convert the inequality to an equality.

$\log (x) = \log (2 x - 1) - \log (3)$

Step 2

Solve the equation.

Simplify the right side.

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

$\log (x) = \log (\frac{2 x - 1}{3})$

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

$x = \frac{2 x - 1}{3}$

Solve for $x$ .

Multiply both sides by $3$ .

$x \cdot 3 = \frac{2 x - 1}{3} \cdot 3$

Simplify.

Simplify the left side.

Move $3$ to the left of $x$ .

$3 x = \frac{2 x - 1}{3} \cdot 3$

Simplify the right side.

Cancel the common factor of $3$ .

Cancel the common factor.

$3 x = \frac{2 x - 1}{3} \cdot 3$

Rewrite the expression.

$3 x = 2 x - 1$

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

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

$3 x - 2 x = - 1$

Subtract $2 x$ from $3 x$ .

$x = - 1$

Step 3

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

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

$\frac{3 x}{2 x - 1} > 0$

Solve for $x$ .

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

$3 x = 0$

$2 x - 1 = 0$

Divide each term in $3 x = 0$ by $3$ and simplify.

Divide each term in $3 x = 0$ by $3$ .

$\frac{3 x}{3} = \frac{0}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 x}{3} = \frac{0}{3}$

Divide $x$ by $1$ .

$x = \frac{0}{3}$

Simplify the right side.

Divide $0$ by $3$ .

$x = 0$

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 $1$ .

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

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

$x = 0$

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

Consolidate the solutions.

$x = 0, \frac{1}{2}$

Find the domain of $\frac{3 x}{2 x - 1}$ .

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

$2 x - 1 = 0$

Solve for $x$ .

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 $1$ .

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

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

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

Use each root to create test intervals.

$x < 0$

$0 < x < \frac{1}{2}$

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

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 $x < 0$ to see if it makes the inequality true.

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

$x = - 2$

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

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

The left side $1.2$ is greater than the right side $0$ , which means that the given statement is always true.

True

Test a value on the interval $0 < x < \frac{1}{2}$ to see if it makes the inequality true.

Choose a value on the interval $0 < x < \frac{1}{2}$ and see if this value makes the original inequality true.

$x = 0.25$

Replace $x$ with $0.25$ in the original inequality.

$\frac{3 (0.25)}{2 (0.25) - 1} > 0$

The left side $- 1.5$ 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.

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

The left side $1.8$ 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.

$x < 0$ True

$0 < x < \frac{1}{2}$ False

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

$x < 0$ True

$0 < x < \frac{1}{2}$ False

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

The solution consists of all of the true intervals.

$x < 0$ or $x > \frac{1}{2}$

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

$2 x - 1 = 0$

Solve for $x$ .

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 $1$ .

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

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

$(- \infty, 0) \cup (\frac{1}{2}, \infty)$

Step 4

The solution consists of all of the true intervals.

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

Step 5

The result can be shown in multiple forms.

Inequality Form:

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

Interval Notation:

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

Step 6

Solve the Inequality for x log of x> = log of 2x-1- log of 3

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