Calculus Examples

$\ln (1 - x) + \ln (2 - x) = \ln (x + 14)$

Step 1

Simplify the left side.

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Step 1.1

Use the product property of logarithms, $\log_{b} (x) + \log_{b} (y) = \log_{b} (x y)$ .

$\ln ((1 - x) (2 - x)) = \ln (x + 14)$

Step 1.2

Expand $(1 - x) (2 - x)$ using the FOIL Method.

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Step 1.2.1

Apply the distributive property.

$\ln (1 (2 - x) - x (2 - x)) = \ln (x + 14)$

Step 1.2.2

Apply the distributive property.

$\ln (1 \cdot 2 + 1 (- x) - x (2 - x)) = \ln (x + 14)$

Step 1.2.3

Apply the distributive property.

$\ln (1 \cdot 2 + 1 (- x) - x \cdot 2 - x (- x)) = \ln (x + 14)$

Step 1.3

Simplify and combine like terms.

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Step 1.3.1

Simplify each term.

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Step 1.3.1.1

Multiply $2$ by $1$ .

$\ln (2 + 1 (- x) - x \cdot 2 - x (- x)) = \ln (x + 14)$

Step 1.3.1.2

Multiply $- x$ by $1$ .

$\ln (2 - x - x \cdot 2 - x (- x)) = \ln (x + 14)$

Step 1.3.1.3

Multiply $2$ by $- 1$ .

$\ln (2 - x - 2 x - x (- x)) = \ln (x + 14)$

Step 1.3.1.4

Rewrite using the commutative property of multiplication.

$\ln (2 - x - 2 x - 1 \cdot (- 1 x \cdot x)) = \ln (x + 14)$

Step 1.3.1.5

Multiply $x$ by $x$ by adding the exponents.

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Step 1.3.1.5.1

Move $x$ .

$\ln (2 - x - 2 x - 1 \cdot (- 1 (x \cdot x))) = \ln (x + 14)$

Step 1.3.1.5.2

Multiply $x$ by $x$ .

$\ln (2 - x - 2 x - 1 \cdot (- 1 x^{2})) = \ln (x + 14)$

Step 1.3.1.6

Multiply $- 1$ by $- 1$ .

$\ln (2 - x - 2 x + 1 x^{2}) = \ln (x + 14)$

Step 1.3.1.7

Multiply $x^{2}$ by $1$ .

$\ln (2 - x - 2 x + x^{2}) = \ln (x + 14)$

Step 1.3.2

Subtract $2 x$ from $- x$ .

$\ln (2 - 3 x + x^{2}) = \ln (x + 14)$

Step 2

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

$2 - 3 x + x^{2} = x + 14$

Step 3

Solve for $x$ .

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Step 3.1

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

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Step 3.1.1

Subtract $x$ from both sides of the equation.

$2 - 3 x + x^{2} - x = 14$

Step 3.1.2

Subtract $x$ from $- 3 x$ .

$2 + x^{2} - 4 x = 14$

Step 3.2

Subtract $14$ from both sides of the equation.

$2 + x^{2} - 4 x - 14 = 0$

Step 3.3

Subtract $14$ from $2$ .

$x^{2} - 4 x - 12 = 0$

Step 3.4

Factor $x^{2} - 4 x - 12$ using the AC method.

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Step 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 $- 12$ and whose sum is $- 4$ .

$- 6, 2$

Step 3.4.2

Write the factored form using these integers.

$(x - 6) (x + 2) = 0$

Step 3.5

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

$x - 6 = 0$

$x + 2 = 0$

Step 3.6

Set $x - 6$ equal to $0$ and solve for $x$ .

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Step 3.6.1

Set $x - 6$ equal to $0$ .

$x - 6 = 0$

Step 3.6.2

Add $6$ to both sides of the equation.

$x = 6$

Step 3.7

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

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Step 3.7.1

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

$x + 2 = 0$

Step 3.7.2

Subtract $2$ from both sides of the equation.

$x = - 2$

Step 3.8

The final solution is all the values that make $(x - 6) (x + 2) = 0$ true.

$x = 6, - 2$

Step 4

Exclude the solutions that do not make $\ln (1 - x) + \ln (2 - x) = \ln (x + 14)$ true.

$x = - 2$