Precalculus Examples

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

Step 1

Write in exponential form.

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Step 1.1

For logarithmic equations, $\log_{b} (x) = y$ 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 1.2

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

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

Step 2

Solve for $x$ .

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Step 2.1

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

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

Step 2.2

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

$| x - 1 | = | 3 |$

Step 2.3

Solve for $x$ .

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Step 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 2.3.2

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

$x - 1 = \pm 3$

Step 2.3.3

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 $\pm$ to find the first solution.

$x - 1 = 3$

Step 2.3.3.2

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

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Step 2.3.3.2.1

Add $1$ to both sides of the equation.

$x = 3 + 1$

Step 2.3.3.2.2

Add $3$ and $1$ .

$x = 4$

Step 2.3.3.3

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

$x - 1 = - 3$

Step 2.3.3.4

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

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Step 2.3.3.4.1

Add $1$ to both sides of the equation.

$x = - 3 + 1$

Step 2.3.3.4.2

Add $- 3$ and $1$ .

$x = - 2$

Step 2.3.3.5

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

$x = 4, - 2$