Precalculus Examples

$\ln (x + 3) = \ln (8) - \ln (x - 4)$

Step 1

Simplify the right side.

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Step 1.1

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

$\ln (x + 3) = \ln (\frac{8}{x - 4})$

Step 2

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

$x + 3 = \frac{8}{x - 4}$

Step 3

Solve for $x$ .

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Step 3.1

Subtract $3$ from both sides of the equation.

$x = \frac{8}{x - 4} - 3$

Step 3.2

Find the LCD of the terms in the equation.

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Step 3.2.1

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

$1, x - 4, 1$

Step 3.2.2

Remove parentheses.

$1, x - 4, 1$

Step 3.2.3

The LCM of one and any expression is the expression.

$x - 4$

Step 3.3

Multiply each term in $x = \frac{8}{x - 4} - 3$ by $x - 4$ to eliminate the fractions.

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Step 3.3.1

Multiply each term in $x = \frac{8}{x - 4} - 3$ by $x - 4$ .

$x (x - 4) = \frac{8}{x - 4} (x - 4) - 3 (x - 4)$

Step 3.3.2

Simplify the left side.

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Step 3.3.2.1

Apply the distributive property.

$x \cdot x + x \cdot - 4 = \frac{8}{x - 4} (x - 4) - 3 (x - 4)$

Step 3.3.2.2

Simplify the expression.

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Step 3.3.2.2.1

Multiply $x$ by $x$ .

$x^{2} + x \cdot - 4 = \frac{8}{x - 4} (x - 4) - 3 (x - 4)$

Step 3.3.2.2.2

Move $- 4$ to the left of $x$ .

$x^{2} - 4 x = \frac{8}{x - 4} (x - 4) - 3 (x - 4)$

Step 3.3.3

Simplify the right side.

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Step 3.3.3.1

Simplify each term.

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Step 3.3.3.1.1

Cancel the common factor of $x - 4$ .

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Step 3.3.3.1.1.1

Cancel the common factor.

$x^{2} - 4 x = \frac{8}{x - 4} (x - 4) - 3 (x - 4)$

Step 3.3.3.1.1.2

Rewrite the expression.

$x^{2} - 4 x = 8 - 3 (x - 4)$

Step 3.3.3.1.2

Apply the distributive property.

$x^{2} - 4 x = 8 - 3 x - 3 \cdot - 4$

Step 3.3.3.1.3

Multiply $- 3$ by $- 4$ .

$x^{2} - 4 x = 8 - 3 x + 12$

Step 3.3.3.2

Add $8$ and $12$ .

$x^{2} - 4 x = - 3 x + 20$

Step 3.4

Solve the equation.

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Step 3.4.1

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

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Step 3.4.1.1

Add $3 x$ to both sides of the equation.

$x^{2} - 4 x + 3 x = 20$

Step 3.4.1.2

Add $- 4 x$ and $3 x$ .

$x^{2} - x = 20$

Step 3.4.2

Subtract $20$ from both sides of the equation.

$x^{2} - x - 20 = 0$

Step 3.4.3

Factor $x^{2} - x - 20$ using the AC method.

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Step 3.4.3.1

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 $- 20$ and whose sum is $- 1$ .

$- 5, 4$

Step 3.4.3.2

Write the factored form using these integers.

$(x - 5) (x + 4) = 0$

Step 3.4.4

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

$x - 5 = 0$

$x + 4 = 0$

Step 3.4.5

Set $x - 5$ equal to $0$ and solve for $x$ .

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Step 3.4.5.1

Set $x - 5$ equal to $0$ .

$x - 5 = 0$

Step 3.4.5.2

Add $5$ to both sides of the equation.

$x = 5$

Step 3.4.6

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

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Step 3.4.6.1

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

$x + 4 = 0$

Step 3.4.6.2

Subtract $4$ from both sides of the equation.

$x = - 4$

Step 3.4.7

The final solution is all the values that make $(x - 5) (x + 4) = 0$ true.

$x = 5, - 4$

Step 4

Exclude the solutions that do not make $\ln (x + 3) = \ln (8) - \ln (x - 4)$ true.

$x = 5$