Precalculus Examples

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

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

Step 1

Write in exponential form.

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Step 1.1

For logarithmic equations,

\log_{b} (x) = y

$\log_{b} (x) = y$ is equivalent to

b^{y} = x

$b^{y} = x$ such that

x > 0

$x > 0$ ,

b > 0

$b > 0$ , and

b \neq 1

$b \neq 1$ . In this case,

b = 3

$b = 3$ ,

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

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

y = 2

$y = 2$ .

b = 3

$b = 3$

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

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

y = 2

$y = 2$

Step 1.2

Substitute the values of

b

$b$ ,

x

$x$ , and

y

$y$ into the equation

b^{y} = x

$b^{y} = x$ .

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

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

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

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

Step 2

Solve for

x

$x$ .

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Step 2.1

Rewrite the equation as

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

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

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

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

Step 2.2

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

| x - 1 | = | 3 |

$| x - 1 | = | 3 |$

Step 2.3

Solve for

x

$x$ .

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Step 2.3.1

Remove the absolute value term. This creates a

\pm

$\pm$ on the right side of the equation because

| x | = \pm x

$| x | = \pm x$ .

x - 1 = \pm | 3 |

$x - 1 = \pm | 3 |$

Step 2.3.2

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

0

$0$ and

3

$3$ is

3

$3$ .

x - 1 = \pm 3

$x - 1 = \pm 3$

Step 2.3.3

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

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Step 2.3.3.1

First, use the positive value of the

\pm

$\pm$ to find the first solution.

x - 1 = 3

$x - 1 = 3$

Step 2.3.3.2

Move all terms not containing

x

$x$ to the right side of the equation.

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Step 2.3.3.2.1

Add

1

$1$ to both sides of the equation.

x = 3 + 1

$x = 3 + 1$

Step 2.3.3.2.2

Add

3

$3$ and

1

$1$ .

x = 4

$x = 4$

x = 4

$x = 4$

Step 2.3.3.3

Next, use the negative value of the

\pm

$\pm$ to find the second solution.

x - 1 = - 3

$x - 1 = - 3$

Step 2.3.3.4

Move all terms not containing

x

$x$ to the right side of the equation.

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Step 2.3.3.4.1

Add

1

$1$ to both sides of the equation.

x = - 3 + 1

$x = - 3 + 1$

Step 2.3.3.4.2

Add

- 3

$- 3$ and

1

$1$ .

x = - 2

$x = - 2$

x = - 2

$x = - 2$

Step 2.3.3.5

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

x = 4, - 2

$x = 4, - 2$

x = 4, - 2

$x = 4, - 2$

x = 4, - 2

$x = 4, - 2$

x = 4, - 2

$x = 4, - 2$