Precalculus Examples

$\log_{6} (x + 4) + \log_{6} (x + 1) = 1$

Step 1

Simplify the left side.

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Step 1.1

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

$\log_{6} ((x + 4) (x + 1)) = 1$

Step 1.2

Expand $(x + 4) (x + 1)$ using the FOIL Method.

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Step 1.2.1

Apply the distributive property.

$\log_{6} (x (x + 1) + 4 (x + 1)) = 1$

Step 1.2.2

Apply the distributive property.

$\log_{6} (x \cdot x + x \cdot 1 + 4 (x + 1)) = 1$

Step 1.2.3

Apply the distributive property.

$\log_{6} (x \cdot x + x \cdot 1 + 4 x + 4 \cdot 1) = 1$

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 $x$ by $x$ .

$\log_{6} (x^{2} + x \cdot 1 + 4 x + 4 \cdot 1) = 1$

Step 1.3.1.2

Multiply $x$ by $1$ .

$\log_{6} (x^{2} + x + 4 x + 4 \cdot 1) = 1$

Step 1.3.1.3

Multiply $4$ by $1$ .

$\log_{6} (x^{2} + x + 4 x + 4) = 1$

Step 1.3.2

Add $x$ and $4 x$ .

$\log_{6} (x^{2} + 5 x + 4) = 1$

Step 2

Rewrite $\log_{6} (x^{2} + 5 x + 4) = 1$ in exponential form using the definition of a logarithm. If $x$ and $b$ are positive real numbers and $b \neq 1$ , then $\log_{b} (x) = y$ is equivalent to $b^{y} = x$ .

$6^{1} = x^{2} + 5 x + 4$

Step 3

Solve for $x$ .

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Step 3.1

Rewrite the equation as $x^{2} + 5 x + 4 = 6^{1}$ .

$x^{2} + 5 x + 4 = 6$

Step 3.2

Move all terms to the left side of the equation and simplify.

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Step 3.2.1

Subtract $6$ from both sides of the equation.

$x^{2} + 5 x + 4 - 6 = 0$

Step 3.2.2

Subtract $6$ from $4$ .

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

Step 3.3

Use the quadratic formula to find the solutions.

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

Step 3.4

Substitute the values $a = 1$ , $b = 5$ , and $c = - 2$ into the quadratic formula and solve for $x$ .

$\frac{- 5 \pm \sqrt{5^{2} - 4 \cdot (1 \cdot - 2)}}{2 \cdot 1}$

Step 3.5

Simplify.

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Step 3.5.1

Simplify the numerator.

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Step 3.5.1.1

Raise $5$ to the power of $2$ .

$x = \frac{- 5 \pm \sqrt{25 - 4 \cdot 1 \cdot - 2}}{2 \cdot 1}$

Step 3.5.1.2

Multiply $- 4 \cdot 1 \cdot - 2$ .

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Step 3.5.1.2.1

Multiply $- 4$ by $1$ .

$x = \frac{- 5 \pm \sqrt{25 - 4 \cdot - 2}}{2 \cdot 1}$

Step 3.5.1.2.2

Multiply $- 4$ by $- 2$ .

$x = \frac{- 5 \pm \sqrt{25 + 8}}{2 \cdot 1}$

Step 3.5.1.3

Add $25$ and $8$ .

$x = \frac{- 5 \pm \sqrt{33}}{2 \cdot 1}$

Step 3.5.2

Multiply $2$ by $1$ .

$x = \frac{- 5 \pm \sqrt{33}}{2}$

Step 3.6

Simplify the expression to solve for the $+$ portion of the $\pm$ .

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Step 3.6.1

Simplify the numerator.

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Step 3.6.1.1

Raise $5$ to the power of $2$ .

$x = \frac{- 5 \pm \sqrt{25 - 4 \cdot 1 \cdot - 2}}{2 \cdot 1}$

Step 3.6.1.2

Multiply $- 4 \cdot 1 \cdot - 2$ .

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Step 3.6.1.2.1

Multiply $- 4$ by $1$ .

$x = \frac{- 5 \pm \sqrt{25 - 4 \cdot - 2}}{2 \cdot 1}$

Step 3.6.1.2.2

Multiply $- 4$ by $- 2$ .

$x = \frac{- 5 \pm \sqrt{25 + 8}}{2 \cdot 1}$

Step 3.6.1.3

Add $25$ and $8$ .

$x = \frac{- 5 \pm \sqrt{33}}{2 \cdot 1}$

Step 3.6.2

Multiply $2$ by $1$ .

$x = \frac{- 5 \pm \sqrt{33}}{2}$

Step 3.6.3

Change the $\pm$ to $+$ .

$x = \frac{- 5 + \sqrt{33}}{2}$

Step 3.6.4

Rewrite $- 5$ as $- 1 (5)$ .

$x = \frac{- 1 \cdot 5 + \sqrt{33}}{2}$

Step 3.6.5

Factor $- 1$ out of $\sqrt{33}$ .

$x = \frac{- 1 \cdot 5 - 1 (- \sqrt{33})}{2}$

Step 3.6.6

Factor $- 1$ out of $- 1 (5) - 1 (- \sqrt{33})$ .

$x = \frac{- 1 (5 - \sqrt{33})}{2}$

Step 3.6.7

Move the negative in front of the fraction.

$x = - \frac{5 - \sqrt{33}}{2}$

Step 3.7

Simplify the expression to solve for the $-$ portion of the $\pm$ .

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Step 3.7.1

Simplify the numerator.

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Step 3.7.1.1

Raise $5$ to the power of $2$ .

$x = \frac{- 5 \pm \sqrt{25 - 4 \cdot 1 \cdot - 2}}{2 \cdot 1}$

Step 3.7.1.2

Multiply $- 4 \cdot 1 \cdot - 2$ .

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Step 3.7.1.2.1

Multiply $- 4$ by $1$ .

$x = \frac{- 5 \pm \sqrt{25 - 4 \cdot - 2}}{2 \cdot 1}$

Step 3.7.1.2.2

Multiply $- 4$ by $- 2$ .

$x = \frac{- 5 \pm \sqrt{25 + 8}}{2 \cdot 1}$

Step 3.7.1.3

Add $25$ and $8$ .

$x = \frac{- 5 \pm \sqrt{33}}{2 \cdot 1}$

Step 3.7.2

Multiply $2$ by $1$ .

$x = \frac{- 5 \pm \sqrt{33}}{2}$

Step 3.7.3

Change the $\pm$ to $-$ .

$x = \frac{- 5 - \sqrt{33}}{2}$

Step 3.7.4

Rewrite $- 5$ as $- 1 (5)$ .

$x = \frac{- 1 \cdot 5 - \sqrt{33}}{2}$

Step 3.7.5

Factor $- 1$ out of $- \sqrt{33}$ .

$x = \frac{- 1 \cdot 5 - (\sqrt{33})}{2}$

Step 3.7.6

Factor $- 1$ out of $- 1 (5) - (\sqrt{33})$ .

$x = \frac{- 1 (5 + \sqrt{33})}{2}$

Step 3.7.7

Move the negative in front of the fraction.

$x = - \frac{5 + \sqrt{33}}{2}$

Step 3.8

The final answer is the combination of both solutions.

$x = - \frac{5 - \sqrt{33}}{2}, - \frac{5 + \sqrt{33}}{2}$

Step 4

Exclude the solutions that do not make $\log_{6} (x + 4) + \log_{6} (x + 1) = 1$ true.

$x = - \frac{5 - \sqrt{33}}{2}$

Step 5

The result can be shown in multiple forms.

Exact Form:

$x = - \frac{5 - \sqrt{33}}{2}$

Decimal Form:

$x = 0.37228132 \dots$