Precalculus Examples

$\ln (x) + \ln (x + 6) = \frac{1}{2} \cdot \ln (9)$

Step 1

Simplify the left side.

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Step 1.1

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

$\ln (x (x + 6)) = \frac{1}{2} \cdot \ln (9)$

Step 1.2

Apply the distributive property.

$\ln (x \cdot x + x \cdot 6) = \frac{1}{2} \cdot \ln (9)$

Step 1.3

Simplify the expression.

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Step 1.3.1

Multiply $x$ by $x$ .

$\ln (x^{2} + x \cdot 6) = \frac{1}{2} \cdot \ln (9)$

Step 1.3.2

Move $6$ to the left of $x$ .

$\ln (x^{2} + 6 x) = \frac{1}{2} \cdot \ln (9)$

Step 2

Simplify the right side.

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Step 2.1

Combine $\frac{1}{2}$ and $\ln (9)$ .

$\ln (x^{2} + 6 x) = \frac{\ln (9)}{2}$

Step 2.2

Rewrite $\ln (9)$ as $\ln (3^{2})$ .

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

Step 2.3

Expand $\ln (3^{2})$ by moving $2$ outside the logarithm.

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

Step 2.4

Cancel the common factor of $2$ .

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Step 2.4.1

Cancel the common factor.

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

Step 2.4.2

Divide $\ln (3)$ by $1$ .

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

Step 3

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

$x^{2} + 6 x = 3$

Step 4

Solve for $x$ .

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Step 4.1

Subtract $3$ from both sides of the equation.

$x^{2} + 6 x - 3 = 0$

Step 4.2

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Step 4.3

Substitute the values $a = 1$ , $b = 6$ , and $c = - 3$ into the quadratic formula and solve for $x$ .

$\frac{- 6 \pm \sqrt{6^{2} - 4 \cdot (1 \cdot - 3)}}{2 \cdot 1}$

Step 4.4

Simplify.

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Step 4.4.1

Simplify the numerator.

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Step 4.4.1.1

Raise $6$ to the power of $2$ .

$x = \frac{- 6 \pm \sqrt{36 - 4 \cdot 1 \cdot - 3}}{2 \cdot 1}$

Step 4.4.1.2

Multiply $- 4 \cdot 1 \cdot - 3$ .

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Step 4.4.1.2.1

Multiply $- 4$ by $1$ .

$x = \frac{- 6 \pm \sqrt{36 - 4 \cdot - 3}}{2 \cdot 1}$

Step 4.4.1.2.2

Multiply $- 4$ by $- 3$ .

$x = \frac{- 6 \pm \sqrt{36 + 12}}{2 \cdot 1}$

Step 4.4.1.3

Add $36$ and $12$ .

$x = \frac{- 6 \pm \sqrt{48}}{2 \cdot 1}$

Step 4.4.1.4

Rewrite $48$ as $4^{2} \cdot 3$ .

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Step 4.4.1.4.1

Factor $16$ out of $48$ .

$x = \frac{- 6 \pm \sqrt{16 (3)}}{2 \cdot 1}$

Step 4.4.1.4.2

Rewrite $16$ as $4^{2}$ .

$x = \frac{- 6 \pm \sqrt{4^{2} \cdot 3}}{2 \cdot 1}$

Step 4.4.1.5

Pull terms out from under the radical.

$x = \frac{- 6 \pm 4 \sqrt{3}}{2 \cdot 1}$

Step 4.4.2

Multiply $2$ by $1$ .

$x = \frac{- 6 \pm 4 \sqrt{3}}{2}$

Step 4.4.3

Simplify $\frac{- 6 \pm 4 \sqrt{3}}{2}$ .

$x = - 3 \pm 2 \sqrt{3}$

Step 4.5

The final answer is the combination of both solutions.

$x = - 3 + 2 \sqrt{3}, - 3 - 2 \sqrt{3}$

Step 5

Exclude the solutions that do not make $\ln (x) + \ln (x + 6) = \frac{1}{2} \cdot \ln (9)$ true.

$x = - 3 + 2 \sqrt{3}$

Step 6

The result can be shown in multiple forms.

Exact Form:

$x = - 3 + 2 \sqrt{3}$

Decimal Form:

$x = 0.46410161 \dots$