Solve for x log base x-3 of 36>2

$\log_{x - 3} (36) > 2$

Step 1

Convert the inequality to an equality.

$\log_{x - 3} (36) = 2$

Step 2

Solve the equation.

Rewrite $\log_{x - 3} (36) = 2$ in exponential form using the definition of a logarithm. If $x$ and $b$ are positive real numbers and $b \neq 1$ , then $\log_{b} (x) = y$ is equivalent to $b^{y} = x$ .

${(x - 3)}^{2} = 36$

Solve for $x$ .

Take the square root of both sides of the equation to eliminate the exponent on the left side.

$x - 3 = \pm \sqrt{36}$

Simplify $\pm \sqrt{36}$ .

Rewrite $36$ as $6^{2}$ .

$x - 3 = \pm \sqrt{6^{2}}$

Pull terms out from under the radical, assuming positive real numbers.

$x - 3 = \pm 6$

The complete solution is the result of both the positive and negative portions of the solution.

First, use the positive value of the $\pm$ to find the first solution.

$x - 3 = 6$

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

Add $3$ to both sides of the equation.

$x = 6 + 3$

Add $6$ and $3$ .

$x = 9$

Next, use the negative value of the $\pm$ to find the second solution.

$x - 3 = - 6$

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

Add $3$ to both sides of the equation.

$x = - 6 + 3$

Add $- 6$ and $3$ .

$x = - 3$

The complete solution is the result of both the positive and negative portions of the solution.

$x = 9, - 3$

Step 3

Find the domain of $\log_{x - 3} (36) - 2$ .

Set the base in $\log_{x - 3} (36)$ greater than $0$ to find where the expression is defined.

$x - 3 > 0$

Add $3$ to both sides of the inequality.

$x > 3$

Set the base in $\log_{x - 3} (36)$ equal to $1$ to find where the expression is undefined.

$x - 3 = 1$

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

Add $3$ to both sides of the equation.

$x = 1 + 3$

Add $1$ and $3$ .

$x = 4$

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

$(3, 4) \cup (4, \infty)$

Step 4

Use each root to create test intervals.

$x < - 3$

$- 3 < x < 3$

$3 < x < 4$

$4 < x < 9$

$x > 9$

Step 5

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 $x < - 3$ to see if it makes the inequality true.

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

$x = - 6$

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

$\log_{(- 6) - 3} (36) > 2$

The logarithm of a negative number is undefined.

Undefined

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

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

$x = 0$

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

$\log_{(0) - 3} (36) > 2$

The logarithm of a negative number is undefined.

Undefined

Test a value on the interval $3 < x < 4$ to see if it makes the inequality true.

Choose a value on the interval $3 < x < 4$ and see if this value makes the original inequality true.

$x = 3.5$

Replace $x$ with $3.5$ in the original inequality.

$\log_{(3.5) - 3} (36) > 2$

The left side $- 5.169925$ is not greater than the right side $2$ , which means that the given statement is false.

False

Test a value on the interval $4 < x < 9$ to see if it makes the inequality true.

Choose a value on the interval $4 < x < 9$ and see if this value makes the original inequality true.

$x = 6$

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

$\log_{(6) - 3} (36) > 2$

The left side $3.2618595$ is greater than the right side $2$ , which means that the given statement is always true.

True

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.

$\log_{(12) - 3} (36) > 2$

The left side $1.63092975$ is not greater than the right side $2$ , which means that the given statement is false.

False

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

$x < - 3$ Undefined

$- 3 < x < 3$ Undefined

$3 < x < 4$ False

$4 < x < 9$ True

$x > 9$ False

Undefined

Step 6

The solution consists of all of the true intervals.

$4 < x < 9$

Step 7

The result can be shown in multiple forms.

Inequality Form:

$4 < x < 9$

Interval Notation:

$(4, 9)$

Step 8

Solve for x log base x-3 of 36>2

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