Solve for x log base 3 of x+12+ log base 3 of x-12=4

$\log_{3} (x + 12) + \log_{3} (x - 12) = 4$

Step 1

Simplify the left side.

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

$\log_{3} ((x + 12) (x - 12)) = 4$

Expand $(x + 12) (x - 12)$ using the FOIL Method.

Apply the distributive property.

$\log_{3} (x (x - 12) + 12 (x - 12)) = 4$

Apply the distributive property.

$\log_{3} (x \cdot x + x \cdot - 12 + 12 (x - 12)) = 4$

Apply the distributive property.

$\log_{3} (x \cdot x + x \cdot - 12 + 12 x + 12 \cdot - 12) = 4$

Simplify terms.

Combine the opposite terms in $x \cdot x + x \cdot - 12 + 12 x + 12 \cdot - 12$ .

Reorder the factors in the terms $x \cdot - 12$ and $12 x$ .

$\log_{3} (x \cdot x - 12 x + 12 x + 12 \cdot - 12) = 4$

Add $- 12 x$ and $12 x$ .

$\log_{3} (x \cdot x + 0 + 12 \cdot - 12) = 4$

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

$\log_{3} (x \cdot x + 12 \cdot - 12) = 4$

Simplify each term.

Multiply $x$ by $x$ .

$\log_{3} (x^{2} + 12 \cdot - 12) = 4$

Multiply $12$ by $- 12$ .

$\log_{3} (x^{2} - 144) = 4$

Step 2

Rewrite $\log_{3} (x^{2} - 144) = 4$ 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 $b^{y} = x$ .

$3^{4} = x^{2} - 144$

Step 3

Solve for $x$ .

Rewrite the equation as $x^{2} - 144 = 3^{4}$ .

$x^{2} - 144 = 3^{4}$

Raise $3$ to the power of $4$ .

$x^{2} - 144 = 81$

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

Add $144$ to both sides of the equation.

$x^{2} = 81 + 144$

Add $81$ and $144$ .

$x^{2} = 225$

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

$x = \pm \sqrt{225}$

Simplify $\pm \sqrt{225}$ .

Rewrite $225$ as $15^{2}$ .

$x = \pm \sqrt{15^{2}}$

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

$x = \pm 15$

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.

$x = 15$

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

$x = - 15$

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

$x = 15, - 15$

Step 4

Exclude the solutions that do not make $\log_{3} (x + 12) + \log_{3} (x - 12) = 4$ true.

$x = 15$

Solve for x log base 3 of x+12+ log base 3 of x-12=4

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