Algebra Examples

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

Step 1

Convert the inequality to an equality.

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

Step 2

Solve the equation.

Tap for more steps...

Step 2.1

Write in exponential form.

Tap for more steps...

Step 2.1.1

For logarithmic equations, $\log_{b} (x) = y$ is equivalent to $b^{y} = x$ such that $x > 0$ , $b > 0$ , and $b \neq 1$ . In this case, $b = 3$ , $x = {(x - 1)}^{2}$ , and $y = 2$ .

$b = 3$

$x = {(x - 1)}^{2}$

$y = 2$

Step 2.1.2

Substitute the values of $b$ , $x$ , and $y$ into the equation $b^{y} = x$ .

$3^{2} = {(x - 1)}^{2}$

Step 2.2

Solve for $x$ .

Tap for more steps...

Step 2.2.1

Rewrite the equation as ${(x - 1)}^{2} = 3^{2}$ .

${(x - 1)}^{2} = 3^{2}$

Step 2.2.2

Since the exponents are equal, the bases of the exponents on both sides of the equation must be equal.

$| x - 1 | = | 3 |$

Step 2.2.3

Solve for $x$ .

Tap for more steps...

Step 2.2.3.1

Remove the absolute value term. This creates a $\pm$ on the right side of the equation because $| x | = \pm x$ .

$x - 1 = \pm | 3 |$

Step 2.2.3.2

The absolute value is the distance between a number and zero. The distance between $0$ and $3$ is $3$ .

$x - 1 = \pm 3$

Step 2.2.3.3

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

Tap for more steps...

Step 2.2.3.3.1

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

$x - 1 = 3$

Step 2.2.3.3.2

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

Tap for more steps...

Step 2.2.3.3.2.1

Add $1$ to both sides of the equation.

$x = 3 + 1$

Step 2.2.3.3.2.2

Add $3$ and $1$ .

$x = 4$

Step 2.2.3.3.3

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

$x - 1 = - 3$

Step 2.2.3.3.4

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

Tap for more steps...

Step 2.2.3.3.4.1

Add $1$ to both sides of the equation.

$x = - 3 + 1$

Step 2.2.3.3.4.2

Add $- 3$ and $1$ .

$x = - 2$

Step 2.2.3.3.5

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

$x = 4, - 2$

Step 3

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

Tap for more steps...

Step 3.1

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

${(x - 1)}^{2} > 0$

Step 3.2

Solve for $x$ .

Tap for more steps...

Step 3.2.1

Take the specified root of both sides of the inequality to eliminate the exponent on the left side.

$\sqrt{{(x - 1)}^{2}} > \sqrt{0}$

Step 3.2.2

Simplify the equation.

Tap for more steps...

Step 3.2.2.1

Simplify the left side.

Tap for more steps...

Step 3.2.2.1.1

Pull terms out from under the radical.

$| x - 1 | > \sqrt{0}$

Step 3.2.2.2

Simplify the right side.

Tap for more steps...

Step 3.2.2.2.1

Simplify $\sqrt{0}$ .

Tap for more steps...

Step 3.2.2.2.1.1

Rewrite $0$ as $0^{2}$ .

$| x - 1 | > \sqrt{0^{2}}$

Step 3.2.2.2.1.2

Pull terms out from under the radical.

$| x - 1 | > | 0 |$

Step 3.2.2.2.1.3

The absolute value is the distance between a number and zero. The distance between $0$ and $0$ is $0$ .

$| x - 1 | > 0$

Step 3.2.3

Write $| x - 1 | > 0$ as a piecewise.

Tap for more steps...

Step 3.2.3.1

To find the interval for the first piece, find where the inside of the absolute value is non-negative.

$x - 1 \geq 0$

Step 3.2.3.2

Add $1$ to both sides of the inequality.

$x \geq 1$

Step 3.2.3.3

In the piece where $x - 1$ is non-negative, remove the absolute value.

$x - 1 > 0$

Step 3.2.3.4

To find the interval for the second piece, find where the inside of the absolute value is negative.

$x - 1 < 0$

Step 3.2.3.5

Add $1$ to both sides of the inequality.

$x < 1$

Step 3.2.3.6

In the piece where $x - 1$ is negative, remove the absolute value and multiply by $- 1$ .

$- (x - 1) > 0$

Step 3.2.3.7

Write as a piecewise.

${\begin{cases} x - 1 > 0 & x \geq 1 \\ - (x - 1) > 0 & x < 1 \end{cases}$

Step 3.2.3.8

Simplify $- (x - 1) > 0$ .

Tap for more steps...

Step 3.2.3.8.1

Apply the distributive property.

${\begin{cases} x - 1 > 0 & x \geq 1 \\ - x - - 1 > 0 & x < 1 \end{cases}$

Step 3.2.3.8.2

Multiply $- 1$ by $- 1$ .

${\begin{cases} x - 1 > 0 & x \geq 1 \\ - x + 1 > 0 & x < 1 \end{cases}$

Step 3.2.4

Add $1$ to both sides of the inequality.

$x > 1$

Step 3.2.5

Solve $- x + 1 > 0$ for $x$ .

Tap for more steps...

Step 3.2.5.1

Subtract $1$ from both sides of the inequality.

$- x > - 1$

Step 3.2.5.2

Divide each term in $- x > - 1$ by $- 1$ and simplify.

Tap for more steps...

Step 3.2.5.2.1

Divide each term in $- x > - 1$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

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

Step 3.2.5.2.2

Simplify the left side.

Tap for more steps...

Step 3.2.5.2.2.1

Dividing two negative values results in a positive value.

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

Step 3.2.5.2.2.2

Divide $x$ by $1$ .

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

Step 3.2.5.2.3

Simplify the right side.

Tap for more steps...

Step 3.2.5.2.3.1

Divide $- 1$ by $- 1$ .

$x < 1$

Step 3.2.6

Find the union of the solutions.

$x < 1$ or $x > 1$

Step 3.3

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

$(- \infty, 1) \cup (1, \infty)$

Step 4

Use each root to create test intervals.

$x < - 2$

$- 2 < x < 1$

$1 < x < 4$

$x > 4$

Step 5

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

Tap for more steps...

Step 5.1

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

Tap for more steps...

Step 5.1.1

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

$x = - 4$

Step 5.1.2

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

$\log_{3} ({((- 4) - 1)}^{2}) > 2$

Step 5.1.3

The left side $2.92994704$ is greater than the right side $2$ , which means that the given statement is always true.

True

Step 5.2

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

Tap for more steps...

Step 5.2.1

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

$x = 0$

Step 5.2.2

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

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

Step 5.2.3

The left side $0$ is not greater than the right side $2$ , which means that the given statement is false.

False

Step 5.3

Test a value on the interval $1 < x < 4$ to see if it makes the inequality true.

Tap for more steps...

Step 5.3.1

Choose a value on the interval $1 < x < 4$ and see if this value makes the original inequality true.

$x = 2$

Step 5.3.2

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

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

Step 5.3.3

The left side $0$ is not greater than the right side $2$ , which means that the given statement is false.

False

Step 5.4

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

Tap for more steps...

Step 5.4.1

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

$x = 6$

Step 5.4.2

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

$\log_{3} ({((6) - 1)}^{2}) > 2$

Step 5.4.3

The left side $2.92994704$ is greater than the right side $2$ , which means that the given statement is always true.

True

Step 5.5

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

$x < - 2$ True

$- 2 < x < 1$ False

$1 < x < 4$ False

$x > 4$ True

$x < - 2$ True

$- 2 < x < 1$ False

$1 < x < 4$ False

$x > 4$ True

Step 6

The solution consists of all of the true intervals.

$x < - 2$ or $x > 4$

Step 7

The result can be shown in multiple forms.

Inequality Form:

$x < - 2 or x > 4$

Interval Notation:

$(- \infty, - 2) \cup (4, \infty)$

Step 8