Solve for x log base 7 of 2x^2-2- log base 7 of x+1=1

$\log_{7} (2 x^{2} - 2) - \log_{7} (x + 1) = 1$

Simplify the left side.

Use the quotient property of logarithms, $\log_{b} (x) - \log_{b} (y) = \log_{b} (\frac{x}{y})$ .

$\log_{7} (\frac{2 x^{2} - 2}{x + 1}) = 1$

Simplify the numerator.

Factor $2$ out of $2 x^{2} - 2$ .

Factor $2$ out of $2 x^{2}$ .

$\log_{7} (\frac{2 (x^{2}) - 2}{x + 1}) = 1$

Factor $2$ out of $- 2$ .

$\log_{7} (\frac{2 (x^{2}) + 2 (- 1)}{x + 1}) = 1$

Factor $2$ out of $2 (x^{2}) + 2 (- 1)$ .

$\log_{7} (\frac{2 (x^{2} - 1)}{x + 1}) = 1$

Rewrite $1$ as $1^{2}$ .

$\log_{7} (\frac{2 (x^{2} - 1^{2})}{x + 1}) = 1$

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x$ and $b = 1$ .

$\log_{7} (\frac{2 (x + 1) (x - 1)}{x + 1}) = 1$

Simplify terms.

Cancel the common factor of $x + 1$ .

Cancel the common factor.

$\log_{7} (\frac{2 (x + 1) (x - 1)}{x + 1}) = 1$

Divide $2 (x - 1)$ by $1$ .

$\log_{7} (2 (x - 1)) = 1$

Apply the distributive property.

$\log_{7} (2 x + 2 \cdot - 1) = 1$

Multiply $2$ by $- 1$ .

$\log_{7} (2 x - 2) = 1$

Rewrite $\log_{7} (2 x - 2) = 1$ 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 $b^{y} = x$ .

$7^{1} = 2 x - 2$

Solve for $x$ .

Rewrite the equation as $2 x - 2 = 7^{1}$ .

$2 x - 2 = 7$

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

Add $2$ to both sides of the equation.

$2 x = 7 + 2$

Add $7$ and $2$ .

$2 x = 9$

Divide each term in $2 x = 9$ by $2$ and simplify.

Divide each term in $2 x = 9$ by $2$ .

$\frac{2 x}{2} = \frac{9}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x}{2} = \frac{9}{2}$

Divide $x$ by $1$ .

$x = \frac{9}{2}$

The result can be shown in multiple forms.

Exact Form:

$x = \frac{9}{2}$

Decimal Form:

$x = 4.5$

Mixed Number Form:

$x = 4 \frac{1}{2}$

Solve for x log base 7 of 2x^2-2- log base 7 of x+1=1

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