Solve for m log base 2 of 2m> log base 2 of m+5

$\log_{2} (2 m) > \log_{2} (m + 5)$

Step 1

Convert the inequality to an equality.

$\log_{2} (2 m) = \log_{2} (m + 5)$

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.

$2 m = m + 5$

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

Subtract $m$ from both sides of the equation.

$2 m - m = 5$

Subtract $m$ from $2 m$ .

$m = 5$

Step 3

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

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

$\frac{2 m}{m + 5} > 0$

Solve for $m$ .

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

$2 m = 0$

$m + 5 = 0$

Divide each term in $2 m = 0$ by $2$ and simplify.

Divide each term in $2 m = 0$ by $2$ .

$\frac{2 m}{2} = \frac{0}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 m}{2} = \frac{0}{2}$

Divide $m$ by $1$ .

$m = \frac{0}{2}$

Simplify the right side.

Divide $0$ by $2$ .

$m = 0$

Subtract $5$ from both sides of the equation.

$m = - 5$

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

$m = 0$

$m = - 5$

Consolidate the solutions.

$m = 0, - 5$

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

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

$m + 5 = 0$

Subtract $5$ from both sides of the equation.

$m = - 5$

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

$(- \infty, - 5) \cup (- 5, \infty)$

Use each root to create test intervals.

$m < - 5$

$- 5 < m < 0$

$m > 0$

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 $m < - 5$ to see if it makes the inequality true.

Choose a value on the interval $m < - 5$ and see if this value makes the original inequality true.

$m = - 8$

Replace $m$ with $- 8$ in the original inequality.

$\frac{2 (- 8)}{(- 8) + 5} > 0$

The left side $5.\overline{3}$ is greater than the right side $0$ , which means that the given statement is always true.

True

Test a value on the interval $- 5 < m < 0$ to see if it makes the inequality true.

Choose a value on the interval $- 5 < m < 0$ and see if this value makes the original inequality true.

$m = - 2$

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

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

The left side $- 1.\overline{3}$ is not greater than the right side $0$ , which means that the given statement is false.

False

Test a value on the interval $m > 0$ to see if it makes the inequality true.

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

$m = 2$

Replace $m$ with $2$ in the original inequality.

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

The left side $0.\overline{571428}$ 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.

$m < - 5$ True

$- 5 < m < 0$ False

$m > 0$ True

$m < - 5$ True

$- 5 < m < 0$ False

$m > 0$ True

The solution consists of all of the true intervals.

$m < - 5$ or $m > 0$

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

$m + 5 = 0$

Subtract $5$ from both sides of the equation.

$m = - 5$

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

$(- \infty, - 5) \cup (0, \infty)$

Step 4

The solution consists of all of the true intervals.

$m > 5$

Step 5

The result can be shown in multiple forms.

Inequality Form:

$m > 5$

Interval Notation:

$(5, \infty)$

Step 6

Solve for m log base 2 of 2m> log base 2 of m+5

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