Precalculus Examples

$\log_{5} (x + 4) + \log_{5} (x - 4) = 2$

Step 1

Subtract $2$ from both sides of the equation.

$\log_{5} (x + 4) + \log_{5} (x - 4) - 2 = 0$

Step 2

Simplify $\log_{5} (x + 4) + \log_{5} (x - 4) - 2$ .

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Step 2.1

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

$\log_{5} ((x + 4) (x - 4)) - 2 = 0$

Step 2.2

Simplify each term.

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Step 2.2.1

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

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Step 2.2.1.1

Apply the distributive property.

$\log_{5} (x (x - 4) + 4 (x - 4)) - 2 = 0$

Step 2.2.1.2

Apply the distributive property.

$\log_{5} (x \cdot x + x \cdot - 4 + 4 (x - 4)) - 2 = 0$

Step 2.2.1.3

Apply the distributive property.

$\log_{5} (x \cdot x + x \cdot - 4 + 4 x + 4 \cdot - 4) - 2 = 0$

Step 2.2.2

Combine the opposite terms in $x \cdot x + x \cdot - 4 + 4 x + 4 \cdot - 4$ .

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Step 2.2.2.1

Reorder the factors in the terms $x \cdot - 4$ and $4 x$ .

$\log_{5} (x \cdot x - 4 x + 4 x + 4 \cdot - 4) - 2 = 0$

Step 2.2.2.2

Add $- 4 x$ and $4 x$ .

$\log_{5} (x \cdot x + 0 + 4 \cdot - 4) - 2 = 0$

Step 2.2.2.3

Add $x \cdot x$ and $0$ .

$\log_{5} (x \cdot x + 4 \cdot - 4) - 2 = 0$

Step 2.2.3

Simplify each term.

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Step 2.2.3.1

Multiply $x$ by $x$ .

$\log_{5} (x^{2} + 4 \cdot - 4) - 2 = 0$

Step 2.2.3.2

Multiply $4$ by $- 4$ .

$\log_{5} (x^{2} - 16) - 2 = 0$

Step 3

Add $2$ to both sides of the equation.

$\log_{5} (x^{2} - 16) = 2$

Step 4

Rewrite $\log_{5} (x^{2} - 16) = 2$ 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 $b^{y} = x$ .

$5^{2} = x^{2} - 16$

Step 5

Solve for $x$ .

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Step 5.1

Rewrite the equation as $x^{2} - 16 = 5^{2}$ .

$x^{2} - 16 = 5^{2}$

Step 5.2

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

$x^{2} - 16 = 25$

Step 5.3

Move all terms not containing $x$ to the right side of the equation.

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Step 5.3.1

Add $16$ to both sides of the equation.

$x^{2} = 25 + 16$

Step 5.3.2

Add $25$ and $16$ .

$x^{2} = 41$

Step 5.4

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$x = \pm \sqrt{41}$

Step 5.5

The complete solution is the result of both the positive and negative portions of the solution.

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Step 5.5.1

First, use the positive value of the $\pm$ to find the first solution.

$x = \sqrt{41}$

Step 5.5.2

Next, use the negative value of the $\pm$ to find the second solution.

$x = - \sqrt{41}$

Step 5.5.3

The complete solution is the result of both the positive and negative portions of the solution.

$x = \sqrt{41}, - \sqrt{41}$

Step 6

The result can be shown in multiple forms.

Exact Form:

$x = \sqrt{41}, - \sqrt{41}$

Decimal Form:

$x = 6.40312423 \dots, - 6.40312423 \dots$