Precalculus Examples

$\ln (x + 4) - \ln (x - 2) = \ln (x)$

Step 1

Use the quotient property of logarithms, $\log_{b} (x) - \log_{b} (y) = \log_{b} (\frac{x}{y})$ .

$\ln (\frac{x + 4}{x - 2}) = \ln (x)$

Step 2

For the equation to be equal, the argument of the logarithms on both sides of the equation must be equal.

$\frac{x + 4}{x - 2} = x$

Step 3

Solve for $x$ .

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Step 3.1

Find the LCD of the terms in the equation.

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Step 3.1.1

Finding the LCD of a list of values is the same as finding the LCM of the denominators of those values.

$x - 2, 1$

Step 3.1.2

Remove parentheses.

$x - 2, 1$

Step 3.1.3

The LCM of one and any expression is the expression.

$x - 2$

Step 3.2

Multiply each term in $\frac{x + 4}{x - 2} = x$ by $x - 2$ to eliminate the fractions.

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Step 3.2.1

Multiply each term in $\frac{x + 4}{x - 2} = x$ by $x - 2$ .

$\frac{x + 4}{x - 2} (x - 2) = x (x - 2)$

Step 3.2.2

Simplify the left side.

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Step 3.2.2.1

Cancel the common factor of $x - 2$ .

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Step 3.2.2.1.1

Cancel the common factor.

$\frac{x + 4}{x - 2} (x - 2) = x (x - 2)$

Step 3.2.2.1.2

Rewrite the expression.

$x + 4 = x (x - 2)$

Step 3.2.3

Simplify the right side.

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Step 3.2.3.1

Apply the distributive property.

$x + 4 = x \cdot x + x \cdot - 2$

Step 3.2.3.2

Simplify the expression.

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Step 3.2.3.2.1

Multiply $x$ by $x$ .

$x + 4 = x^{2} + x \cdot - 2$

Step 3.2.3.2.2

Move $- 2$ to the left of $x$ .

$x + 4 = x^{2} - 2 x$

Step 3.3

Solve the equation.

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Step 3.3.1

Since $x$ is on the right side of the equation, switch the sides so it is on the left side of the equation.

$x^{2} - 2 x = x + 4$

Step 3.3.2

Move all terms containing $x$ to the left side of the equation.

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Step 3.3.2.1

Subtract $x$ from both sides of the equation.

$x^{2} - 2 x - x = 4$

Step 3.3.2.2

Subtract $x$ from $- 2 x$ .

$x^{2} - 3 x = 4$

Step 3.3.3

Subtract $4$ from both sides of the equation.

$x^{2} - 3 x - 4 = 0$

Step 3.3.4

Factor $x^{2} - 3 x - 4$ using the AC method.

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Step 3.3.4.1

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 4$ and whose sum is $- 3$ .

$- 4, 1$

Step 3.3.4.2

Write the factored form using these integers.

$(x - 4) (x + 1) = 0$

Step 3.3.5

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x - 4 = 0$

$x + 1 = 0$

Step 3.3.6

Set $x - 4$ equal to $0$ and solve for $x$ .

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Step 3.3.6.1

Set $x - 4$ equal to $0$ .

$x - 4 = 0$

Step 3.3.6.2

Add $4$ to both sides of the equation.

$x = 4$

Step 3.3.7

Set $x + 1$ equal to $0$ and solve for $x$ .

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Step 3.3.7.1

Set $x + 1$ equal to $0$ .

$x + 1 = 0$

Step 3.3.7.2

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 3.3.8

The final solution is all the values that make $(x - 4) (x + 1) = 0$ true.

$x = 4, - 1$

Step 4

Exclude the solutions that do not make $\ln (x + 4) - \ln (x - 2) = \ln (x)$ true.

$x = 4$