Solve for x log base 2 of x+3+ log base 2 of x-3=4

$\log_{2} (x + 3) + \log_{2} (x - 3) = 4$

Step 1

Simplify the left side.

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

$\log_{2} ((x + 3) (x - 3)) = 4$

Expand $(x + 3) (x - 3)$ using the FOIL Method.

Apply the distributive property.

$\log_{2} (x (x - 3) + 3 (x - 3)) = 4$

Apply the distributive property.

$\log_{2} (x \cdot x + x \cdot - 3 + 3 (x - 3)) = 4$

Apply the distributive property.

$\log_{2} (x \cdot x + x \cdot - 3 + 3 x + 3 \cdot - 3) = 4$

Simplify terms.

Combine the opposite terms in $x \cdot x + x \cdot - 3 + 3 x + 3 \cdot - 3$ .

Reorder the factors in the terms $x \cdot - 3$ and $3 x$ .

$\log_{2} (x \cdot x - 3 x + 3 x + 3 \cdot - 3) = 4$

Add $- 3 x$ and $3 x$ .

$\log_{2} (x \cdot x + 0 + 3 \cdot - 3) = 4$

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

$\log_{2} (x \cdot x + 3 \cdot - 3) = 4$

Simplify each term.

Multiply $x$ by $x$ .

$\log_{2} (x^{2} + 3 \cdot - 3) = 4$

Multiply $3$ by $- 3$ .

$\log_{2} (x^{2} - 9) = 4$

Step 2

Rewrite $\log_{2} (x^{2} - 9) = 4$ 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 $b^{y} = x$ .

$2^{4} = x^{2} - 9$

Step 3

Solve for $x$ .

Rewrite the equation as $x^{2} - 9 = 2^{4}$ .

$x^{2} - 9 = 2^{4}$

Raise $2$ to the power of $4$ .

$x^{2} - 9 = 16$

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

Add $9$ to both sides of the equation.

$x^{2} = 16 + 9$

Add $16$ and $9$ .

$x^{2} = 25$

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

$x = \pm \sqrt{25}$

Simplify $\pm \sqrt{25}$ .

Rewrite $25$ as $5^{2}$ .

$x = \pm \sqrt{5^{2}}$

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

$x = \pm 5$

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.

$x = 5$

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

$x = - 5$

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

$x = 5, - 5$

Step 4

Exclude the solutions that do not make $\log_{2} (x + 3) + \log_{2} (x - 3) = 4$ true.

$x = 5$

Solve for x log base 2 of x+3+ log base 2 of x-3=4

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