Precalculus Examples

$\ln (x) + \ln (x + 5) = \ln (14)$

Step 1

Simplify the left side.

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Step 1.1

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

$\ln (x (x + 5)) = \ln (14)$

Step 1.2

Apply the distributive property.

$\ln (x \cdot x + x \cdot 5) = \ln (14)$

Step 1.3

Simplify the expression.

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Step 1.3.1

Multiply $x$ by $x$ .

$\ln (x^{2} + x \cdot 5) = \ln (14)$

Step 1.3.2

Move $5$ to the left of $x$ .

$\ln (x^{2} + 5 x) = \ln (14)$

Step 2

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

$x^{2} + 5 x = 14$

Step 3

Solve for $x$ .

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Step 3.1

Subtract $14$ from both sides of the equation.

$x^{2} + 5 x - 14 = 0$

Step 3.2

Factor $x^{2} + 5 x - 14$ using the AC method.

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Step 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 $- 14$ and whose sum is $5$ .

$- 2, 7$

Step 3.2.2

Write the factored form using these integers.

$(x - 2) (x + 7) = 0$

Step 3.3

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

$x - 2 = 0$

$x + 7 = 0$

Step 3.4

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

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Step 3.4.1

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

$x - 2 = 0$

Step 3.4.2

Add $2$ to both sides of the equation.

$x = 2$

Step 3.5

Set $x + 7$ equal to $0$ and solve for $x$ .

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Step 3.5.1

Set $x + 7$ equal to $0$ .

$x + 7 = 0$

Step 3.5.2

Subtract $7$ from both sides of the equation.

$x = - 7$

Step 3.6

The final solution is all the values that make $(x - 2) (x + 7) = 0$ true.

$x = 2, - 7$

Step 4

Exclude the solutions that do not make $\ln (x) + \ln (x + 5) = \ln (14)$ true.

$x = 2$