Precalculus Examples

$2 \ln (x) = \ln (3) + \ln (x + 6)$

Step 1

Simplify the left side.

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Step 1.1

Simplify $2 \ln (x)$ by moving $2$ inside the logarithm.

$\ln (x^{2}) = \ln (3) + \ln (x + 6)$

Step 2

Simplify the right side.

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Step 2.1

Simplify $\ln (3) + \ln (x + 6)$ .

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Step 2.1.1

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

$\ln (x^{2}) = \ln (3 (x + 6))$

Step 2.1.2

Apply the distributive property.

$\ln (x^{2}) = \ln (3 x + 3 \cdot 6)$

Step 2.1.3

Multiply $3$ by $6$ .

$\ln (x^{2}) = \ln (3 x + 18)$

Step 3

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

$x^{2} = 3 x + 18$

Step 4

Solve for $x$ .

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Step 4.1

Subtract $3 x$ from both sides of the equation.

$x^{2} - 3 x = 18$

Step 4.2

Subtract $18$ from both sides of the equation.

$x^{2} - 3 x - 18 = 0$

Step 4.3

Factor $x^{2} - 3 x - 18$ using the AC method.

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Step 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 $- 18$ and whose sum is $- 3$ .

$- 6, 3$

Step 4.3.2

Write the factored form using these integers.

$(x - 6) (x + 3) = 0$

Step 4.4

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

$x - 6 = 0$

$x + 3 = 0$

Step 4.5

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

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Step 4.5.1

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

$x - 6 = 0$

Step 4.5.2

Add $6$ to both sides of the equation.

$x = 6$

Step 4.6

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

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Step 4.6.1

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

$x + 3 = 0$

Step 4.6.2

Subtract $3$ from both sides of the equation.

$x = - 3$

Step 4.7

The final solution is all the values that make $(x - 6) (x + 3) = 0$ true.

$x = 6, - 3$

Step 5

Exclude the solutions that do not make $2 \ln (x) = \ln (3) + \ln (x + 6)$ true.

$x = 6$