Solve for x log base 8 of x+1+ log base 8 of x-1=1

$\log_{8} (x + 1) + \log_{8} (x - 1) = 1$

Simplify the left side.

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

$\log_{8} ((x + 1) (x - 1)) = 1$

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

Apply the distributive property.

$\log_{8} (x (x - 1) + 1 (x - 1)) = 1$

Apply the distributive property.

$\log_{8} (x \cdot x + x \cdot - 1 + 1 (x - 1)) = 1$

Apply the distributive property.

$\log_{8} (x \cdot x + x \cdot - 1 + 1 x + 1 \cdot - 1) = 1$

Simplify and combine like terms.

Simplify each term.

Multiply $x$ by $x$ .

$\log_{8} (x^{2} + x \cdot - 1 + 1 x + 1 \cdot - 1) = 1$

Move $- 1$ to the left of $x$ .

$\log_{8} (x^{2} - 1 \cdot x + 1 x + 1 \cdot - 1) = 1$

Rewrite $- 1 x$ as $- x$ .

$\log_{8} (x^{2} - x + 1 x + 1 \cdot - 1) = 1$

Multiply $x$ by $1$ .

$\log_{8} (x^{2} - x + x + 1 \cdot - 1) = 1$

Multiply $- 1$ by $1$ .

$\log_{8} (x^{2} - x + x - 1) = 1$

Add $- x$ and $x$ .

$\log_{8} (x^{2} + 0 - 1) = 1$

Add $x^{2}$ and $0$ .

$\log_{8} (x^{2} - 1) = 1$

Rewrite $\log_{8} (x^{2} - 1) = 1$ 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 $b^{y} = x$ .

$8^{1} = x^{2} - 1$

Solve for $x$ .

Rewrite the equation as $x^{2} - 1 = 8^{1}$ .

$x^{2} - 1 = 8$

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

Add $1$ to both sides of the equation.

$x^{2} = 8 + 1$

Add $8$ and $1$ .

$x^{2} = 9$

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

$x = \pm \sqrt{9}$

Simplify $\pm \sqrt{9}$ .

Rewrite $9$ as $3^{2}$ .

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

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

$x = \pm 3$

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.

$x = 3$

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

$x = - 3$

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

$x = 3, - 3$

Exclude the solutions that do not make $\log_{8} (x + 1) + \log_{8} (x - 1) = 1$ true.

$x = 3$

Solve for x log base 8 of x+1+ log base 8 of x-1=1

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