Precalculus Examples

$\ln (x) = \ln (20) - \ln (x + 1)$

Step 1

Simplify the right side.

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Step 1.1

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

$\ln (x) = \ln (\frac{20}{x + 1})$

Step 2

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

$x = \frac{20}{x + 1}$

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.

$1, x + 1$

Step 3.1.2

Remove parentheses.

$1, x + 1$

Step 3.1.3

The LCM of one and any expression is the expression.

$x + 1$

Step 3.2

Multiply each term in $x = \frac{20}{x + 1}$ by $x + 1$ to eliminate the fractions.

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Step 3.2.1

Multiply each term in $x = \frac{20}{x + 1}$ by $x + 1$ .

$x (x + 1) = \frac{20}{x + 1} (x + 1)$

Step 3.2.2

Simplify the left side.

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Step 3.2.2.1

Apply the distributive property.

$x \cdot x + x \cdot 1 = \frac{20}{x + 1} (x + 1)$

Step 3.2.2.2

Simplify the expression.

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Step 3.2.2.2.1

Multiply $x$ by $x$ .

$x^{2} + x \cdot 1 = \frac{20}{x + 1} (x + 1)$

Step 3.2.2.2.2

Multiply $x$ by $1$ .

$x^{2} + x = \frac{20}{x + 1} (x + 1)$

Step 3.2.3

Simplify the right side.

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Step 3.2.3.1

Cancel the common factor of $x + 1$ .

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Step 3.2.3.1.1

Cancel the common factor.

$x^{2} + x = \frac{20}{x + 1} (x + 1)$

Step 3.2.3.1.2

Rewrite the expression.

$x^{2} + x = 20$

Step 3.3

Solve the equation.

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Step 3.3.1

Subtract $20$ from both sides of the equation.

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

Step 3.3.2

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

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Step 3.3.2.1

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 $1$ .

$- 4, 5$

Step 3.3.2.2

Write the factored form using these integers.

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

Step 3.3.3

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

$x - 4 = 0$

$x + 5 = 0$

Step 3.3.4

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

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Step 3.3.4.1

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

$x - 4 = 0$

Step 3.3.4.2

Add $4$ to both sides of the equation.

$x = 4$

Step 3.3.5

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

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Step 3.3.5.1

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

$x + 5 = 0$

Step 3.3.5.2

Subtract $5$ from both sides of the equation.

$x = - 5$

Step 3.3.6

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

$x = 4, - 5$

Step 4

Exclude the solutions that do not make $\ln (x) = \ln (20) - \ln (x + 1)$ true.

$x = 4$