Precalculus Examples

$y = | x - 2 |$ , $y = \log_{x} (1)$

Step 1

Eliminate the equal sides of each equation and combine.

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

Step 2

Solve $| x - 2 | = \log_{x} (1)$ for $x$ .

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Step 2.1

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

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

Step 2.2

Simplify $\pm \log_{x} (1)$ .

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Step 2.2.1

Logarithm base $x$ of $1$ is $0$ .

$x - 2 = \pm 0$

Step 2.2.2

Plus or minus $0$ is $0$ .

$x - 2 = 0$

Step 2.3

Add $2$ to both sides of the equation.

$x = 2$

Step 3

Evaluate $y$ when $x = 2$ .

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Step 3.1

Substitute $2$ for $x$ .

$y = \log_{2} (1)$

Step 3.2

Substitute $2$ for $x$ in $y = \log_{2} (1)$ and solve for $y$ .

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Step 3.2.1

Remove parentheses.

$y = \log_{2} (1)$

Step 3.2.2

Logarithm base $2$ of $1$ is $0$ .

$y = 0$

Step 4

The solution to the system is the complete set of ordered pairs that are valid solutions.

$(2, 0)$

Step 5

The result can be shown in multiple forms.

Point Form:

$(2, 0)$

Equation Form:

$x = 2, y = 0$

Step 6