Solve for x log base 2 of 3x+5> = log base 2 of x-9

$\log_{2} (3 x + 5) \geq \log_{2} (x - 9)$

Step 1

Convert the inequality to an equality.

$\log_{2} (3 x + 5) = \log_{2} (x - 9)$

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.

$3 x + 5 = x - 9$

Solve for $x$ .

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

Subtract $x$ from both sides of the equation.

$3 x + 5 - x = - 9$

Subtract $x$ from $3 x$ .

$2 x + 5 = - 9$

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

Subtract $5$ from both sides of the equation.

$2 x = - 9 - 5$

Subtract $5$ from $- 9$ .

$2 x = - 14$

Divide each term in $2 x = - 14$ by $2$ and simplify.

Divide each term in $2 x = - 14$ by $2$ .

$\frac{2 x}{2} = \frac{- 14}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x}{2} = \frac{- 14}{2}$

Divide $x$ by $1$ .

$x = \frac{- 14}{2}$

Simplify the right side.

Divide $- 14$ by $2$ .

$x = - 7$

Step 3

Find the domain of $\log_{2} (\frac{3 x + 5}{x - 9})$ .

Set the argument in $\log_{2} (\frac{3 x + 5}{x - 9})$ greater than $0$ to find where the expression is defined.

$\frac{3 x + 5}{x - 9} > 0$

Solve for $x$ .

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

$3 x + 5 = 0$

$x - 9 = 0$

Subtract $5$ from both sides of the equation.

$3 x = - 5$

Divide each term in $3 x = - 5$ by $3$ and simplify.

Divide each term in $3 x = - 5$ by $3$ .

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

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

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

Divide $x$ by $1$ .

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

Simplify the right side.

Move the negative in front of the fraction.

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

Add $9$ to both sides of the equation.

$x = 9$

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

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

$x = 9$

Consolidate the solutions.

$x = - \frac{5}{3}, 9$

Find the domain of $\frac{3 x + 5}{x - 9}$ .

Set the denominator in $\frac{3 x + 5}{x - 9}$ equal to $0$ to find where the expression is undefined.

$x - 9 = 0$

Add $9$ to both sides of the equation.

$x = 9$

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

$(- \infty, 9) \cup (9, \infty)$

Use each root to create test intervals.

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

$- \frac{5}{3} < x < 9$

$x > 9$

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 $x < - \frac{5}{3}$ to see if it makes the inequality true.

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

$x = - 4$

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

$\frac{3 (- 4) + 5}{(- 4) - 9} > 0$

The left side $0.\overline{538461}$ is greater than the right side $0$ , which means that the given statement is always true.

True

Test a value on the interval $- \frac{5}{3} < x < 9$ to see if it makes the inequality true.

Choose a value on the interval $- \frac{5}{3} < x < 9$ and see if this value makes the original inequality true.

$x = 0$

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

$\frac{3 (0) + 5}{(0) - 9} > 0$

The left side $- 0.\overline{5}$ is not greater than the right side $0$ , which means that the given statement is false.

False

Test a value on the interval $x > 9$ to see if it makes the inequality true.

Choose a value on the interval $x > 9$ and see if this value makes the original inequality true.

$x = 12$

Replace $x$ with $12$ in the original inequality.

$\frac{3 (12) + 5}{(12) - 9} > 0$

The left side $13.\overline{6}$ 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.

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

$- \frac{5}{3} < x < 9$ False

$x > 9$ True

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

$- \frac{5}{3} < x < 9$ False

$x > 9$ True

The solution consists of all of the true intervals.

$x < - \frac{5}{3}$ or $x > 9$

Set the denominator in $\frac{3 x + 5}{x - 9}$ equal to $0$ to find where the expression is undefined.

$x - 9 = 0$

Add $9$ to both sides of the equation.

$x = 9$

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

$(- \infty, - \frac{5}{3}) \cup (9, \infty)$

Step 4

The solution consists of all of the true intervals.

$x > 9$

Step 5

The result can be shown in multiple forms.

Inequality Form:

$x > 9$

Interval Notation:

$(9, \infty)$

Step 6

Solve for x log base 2 of 3x+5> = log base 2 of x-9

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