Solve for x log base 6 of x+8+ log base 6 of x-8=2

$\log_{6} (x + 8) + \log_{6} (x - 8) = 2$

Step 1

Simplify the left side.

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

$\log_{6} ((x + 8) (x - 8)) = 2$

Expand $(x + 8) (x - 8)$ using the FOIL Method.

Apply the distributive property.

$\log_{6} (x (x - 8) + 8 (x - 8)) = 2$

Apply the distributive property.

$\log_{6} (x \cdot x + x \cdot - 8 + 8 (x - 8)) = 2$

Apply the distributive property.

$\log_{6} (x \cdot x + x \cdot - 8 + 8 x + 8 \cdot - 8) = 2$

Simplify terms.

Combine the opposite terms in $x \cdot x + x \cdot - 8 + 8 x + 8 \cdot - 8$ .

Reorder the factors in the terms $x \cdot - 8$ and $8 x$ .

$\log_{6} (x \cdot x - 8 x + 8 x + 8 \cdot - 8) = 2$

Add $- 8 x$ and $8 x$ .

$\log_{6} (x \cdot x + 0 + 8 \cdot - 8) = 2$

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

$\log_{6} (x \cdot x + 8 \cdot - 8) = 2$

Simplify each term.

Multiply $x$ by $x$ .

$\log_{6} (x^{2} + 8 \cdot - 8) = 2$

Multiply $8$ by $- 8$ .

$\log_{6} (x^{2} - 64) = 2$

Step 2

Rewrite $\log_{6} (x^{2} - 64) = 2$ 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 $b^{y} = x$ .

$6^{2} = x^{2} - 64$

Step 3

Solve for $x$ .

Rewrite the equation as $x^{2} - 64 = 6^{2}$ .

$x^{2} - 64 = 6^{2}$

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

$x^{2} - 64 = 36$

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

Add $64$ to both sides of the equation.

$x^{2} = 36 + 64$

Add $36$ and $64$ .

$x^{2} = 100$

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

$x = \pm \sqrt{100}$

Simplify $\pm \sqrt{100}$ .

Rewrite $100$ as $10^{2}$ .

$x = \pm \sqrt{10^{2}}$

Pull terms out from under the radical, assuming positive real numbers.

$x = \pm 10$

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

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

$x = 10$

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

$x = - 10$

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

$x = 10, - 10$

Step 4

Exclude the solutions that do not make $\log_{6} (x + 8) + \log_{6} (x - 8) = 2$ true.

$x = 10$

Solve for x log base 6 of x+8+ log base 6 of x-8=2

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