Algebra Examples

$\ln (x + 5) = \ln (x - 1) - \ln (x + 1)$

Step 1

Simplify the right side.

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Step 1.1

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

$\ln (x + 5) = \ln (\frac{x - 1}{x + 1})$

Step 2

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

$x + 5 = \frac{x - 1}{x + 1}$

Step 3

Solve for $x$ .

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Step 3.1

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

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Step 3.1.1

Subtract $5$ from both sides of the equation.

$x = \frac{x - 1}{x + 1} - 5$

Step 3.1.2

Simplify each term.

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Step 3.1.2.1

Split the fraction $\frac{x - 1}{x + 1}$ into two fractions.

$x = \frac{x}{x + 1} + \frac{- 1}{x + 1} - 5$

Step 3.1.2.2

Move the negative in front of the fraction.

$x = \frac{x}{x + 1} - \frac{1}{x + 1} - 5$

Step 3.1.3

Find the common denominator.

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Step 3.1.3.1

Write $- 5$ as a fraction with denominator $1$ .

$x = \frac{x}{x + 1} - \frac{1}{x + 1} + \frac{- 5}{1}$

Step 3.1.3.2

Multiply $\frac{- 5}{1}$ by $\frac{x + 1}{x + 1}$ .

$x = \frac{x}{x + 1} - \frac{1}{x + 1} + \frac{- 5}{1} \cdot \frac{x + 1}{x + 1}$

Step 3.1.3.3

Multiply $\frac{- 5}{1}$ by $\frac{x + 1}{x + 1}$ .

$x = \frac{x}{x + 1} - \frac{1}{x + 1} + \frac{- 5 (x + 1)}{x + 1}$

Step 3.1.4

Combine the numerators over the common denominator.

$x = \frac{x - 1 - 5 (x + 1)}{x + 1}$

Step 3.1.5

Simplify each term.

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Step 3.1.5.1

Apply the distributive property.

$x = \frac{x - 1 - 5 x - 5 \cdot 1}{x + 1}$

Step 3.1.5.2

Multiply $- 5$ by $1$ .

$x = \frac{x - 1 - 5 x - 5}{x + 1}$

Step 3.1.6

Subtract $5 x$ from $x$ .

$x = \frac{- 4 x - 1 - 5}{x + 1}$

Step 3.1.7

Subtract $5$ from $- 1$ .

$x = \frac{- 4 x - 6}{x + 1}$

Step 3.1.8

Factor $2$ out of $- 4 x - 6$ .

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Step 3.1.8.1

Factor $2$ out of $- 4 x$ .

$x = \frac{2 (- 2 x) - 6}{x + 1}$

Step 3.1.8.2

Factor $2$ out of $- 6$ .

$x = \frac{2 (- 2 x) + 2 (- 3)}{x + 1}$

Step 3.1.8.3

Factor $2$ out of $2 (- 2 x) + 2 (- 3)$ .

$x = \frac{2 (- 2 x - 3)}{x + 1}$

Step 3.1.9

Factor $- 1$ out of $- 2 x$ .

$x = \frac{2 (- (2 x) - 3)}{x + 1}$

Step 3.1.10

Rewrite $- 3$ as $- 1 (3)$ .

$x = \frac{2 (- (2 x) - 1 (3))}{x + 1}$

Step 3.1.11

Factor $- 1$ out of $- (2 x) - 1 (3)$ .

$x = \frac{2 (- (2 x + 3))}{x + 1}$

Step 3.1.12

Rewrite $- (2 x + 3)$ as $- 1 (2 x + 3)$ .

$x = \frac{2 (- 1 (2 x + 3))}{x + 1}$

Step 3.1.13

Move the negative in front of the fraction.

$x = - \frac{2 (2 x + 3)}{x + 1}$

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.

$1, x + 1$

Step 3.2.2

Remove parentheses.

$1, x + 1$

Step 3.2.3

The LCM of one and any expression is the expression.

$x + 1$

Step 3.3

Multiply each term in $x = - \frac{2 (2 x + 3)}{x + 1}$ by $x + 1$ to eliminate the fractions.

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Step 3.3.1

Multiply each term in $x = - \frac{2 (2 x + 3)}{x + 1}$ by $x + 1$ .

$x (x + 1) = - \frac{2 (2 x + 3)}{x + 1} (x + 1)$

Step 3.3.2

Simplify the left side.

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Step 3.3.2.1

Apply the distributive property.

$x \cdot x + x \cdot 1 = - \frac{2 (2 x + 3)}{x + 1} (x + 1)$

Step 3.3.2.2

Simplify the expression.

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Step 3.3.2.2.1

Multiply $x$ by $x$ .

$x^{2} + x \cdot 1 = - \frac{2 (2 x + 3)}{x + 1} (x + 1)$

Step 3.3.2.2.2

Multiply $x$ by $1$ .

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

Step 3.3.3

Simplify the right side.

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Step 3.3.3.1

Cancel the common factor of $x + 1$ .

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Step 3.3.3.1.1

Move the leading negative in $- \frac{2 (2 x + 3)}{x + 1}$ into the numerator.

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

Step 3.3.3.1.2

Cancel the common factor.

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

Step 3.3.3.1.3

Rewrite the expression.

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

Step 3.3.3.2

Apply the distributive property.

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

Step 3.3.3.3

Multiply.

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Step 3.3.3.3.1

Multiply $2$ by $- 2$ .

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

Step 3.3.3.3.2

Multiply $- 2$ by $3$ .

$x^{2} + x = - 4 x - 6$

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 $4 x$ to both sides of the equation.

$x^{2} + x + 4 x = - 6$

Step 3.4.1.2

Add $x$ and $4 x$ .

$x^{2} + 5 x = - 6$

Step 3.4.2

Add $6$ to both sides of the equation.

$x^{2} + 5 x + 6 = 0$

Step 3.4.3

Factor $x^{2} + 5 x + 6$ 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 $6$ and whose sum is $5$ .

$2, 3$

Step 3.4.3.2

Write the factored form using these integers.

$(x + 2) (x + 3) = 0$

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 $0$ .

$x + 2 = 0$

$x + 3 = 0$

Step 3.4.5

Set $x + 2$ equal to $0$ and solve for $x$ .

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Step 3.4.5.1

Set $x + 2$ equal to $0$ .

$x + 2 = 0$

Step 3.4.5.2

Subtract $2$ from both sides of the equation.

$x = - 2$

Step 3.4.6

Set $x + 3$ equal to $0$ and solve for $x$ .

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Step 3.4.6.1

Set $x + 3$ equal to $0$ .

$x + 3 = 0$

Step 3.4.6.2

Subtract $3$ from both sides of the equation.

$x = - 3$

Step 3.4.7

The final solution is all the values that make $(x + 2) (x + 3) = 0$ true.

$x = - 2, - 3$

Step 4

Exclude the solutions that do not make $\ln (x + 5) = \ln (x - 1) - \ln (x + 1)$ true.

$No solution$