Solve for x log base 2 of x^2-4- log base 2 of 2x+4=4

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

Simplify the left side.

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

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

Simplify the numerator.

Rewrite $4$ as $2^{2}$ .

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

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 $b = 2$ .

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

Simplify terms.

Factor $2$ out of $2 x + 4$ .

Factor $2$ out of $2 x$ .

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

Factor $2$ out of $4$ .

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

Factor $2$ out of $2 x + 2 \cdot 2$ .

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

Reduce the expression $\frac{(x + 2) (x - 2)}{2 (x + 2)}$ by cancelling the common factors.

Cancel the common factor.

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

Rewrite the expression.

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

Rewrite $\frac{x - 2}{2}$ as $(x - 2) \cdot 2^{- 1}$ .

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

Rewrite $\log_{2} ((x - 2) \cdot 2^{- 1})$ as $\log_{2} (x - 2) + \log_{2} (2^{- 1})$ .

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

Use logarithm rules to move $- 1$ out of the exponent.

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

Logarithm base $2$ of $2$ is $1$ .

$\log_{2} (x - 2) - 1 \cdot 1 = 4$

Multiply $- 1$ by $1$ .

$\log_{2} (x - 2) - 1 = 4$

Move all terms not containing $\log_{2} (x - 2)$ to the right side of the equation.

Add $1$ to both sides of the equation.

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

Add $4$ and $1$ .

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

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

$2^{5} = x - 2$

Solve for $x$ .

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

$x - 2 = 2^{5}$

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

$x - 2 = 32$

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

Add $2$ to both sides of the equation.

$x = 32 + 2$

Add $32$ and $2$ .

$x = 34$

Solve for x log base 2 of x^2-4- log base 2 of 2x+4=4

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