Solve for x log base 3 of x+6+ log base 3 of x-6- log base 3 of x=2

$\log_{3} (x + 6) + \log_{3} (x - 6) - \log_{3} (x) = 2$

Simplify the left side.

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

$\log_{3} ((x + 6) (x - 6)) - \log_{3} (x) = 2$

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

$\log_{3} (\frac{(x + 6) (x - 6)}{x}) = 2$

Rewrite $\log_{3} (\frac{(x + 6) (x - 6)}{x}) = 2$ 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 $b^{y} = x$ .

$3^{2} = \frac{(x + 6) (x - 6)}{x}$

Cross multiply to remove the fraction.

$(x + 6) (x - 6) = 3^{2} (x)$

Raise $3$ to the power of $2$ .

$(x + 6) (x - 6) = 9 x$

Move all terms containing $x$ to the left side of the equation.

Subtract $9 x$ from both sides of the equation.

$(x + 6) (x - 6) - 9 x = 0$

Simplify each term.

Expand $(x + 6) (x - 6)$ using the FOIL Method.

Apply the distributive property.

$x (x - 6) + 6 (x - 6) - 9 x = 0$

Apply the distributive property.

$x \cdot x + x \cdot - 6 + 6 (x - 6) - 9 x = 0$

Apply the distributive property.

$x \cdot x + x \cdot - 6 + 6 x + 6 \cdot - 6 - 9 x = 0$

Combine the opposite terms in $x \cdot x + x \cdot - 6 + 6 x + 6 \cdot - 6$ .

Reorder the factors in the terms $x \cdot - 6$ and $6 x$ .

$x \cdot x - 6 x + 6 x + 6 \cdot - 6 - 9 x = 0$

Add $- 6 x$ and $6 x$ .

$x \cdot x + 0 + 6 \cdot - 6 - 9 x = 0$

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

$x \cdot x + 6 \cdot - 6 - 9 x = 0$

Simplify each term.

Multiply $x$ by $x$ .

$x^{2} + 6 \cdot - 6 - 9 x = 0$

Multiply $6$ by $- 6$ .

$x^{2} - 36 - 9 x = 0$

Factor $x^{2} - 36 - 9 x$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 36$ and whose sum is $- 9$ .

$- 12, 3$

Write the factored form using these integers.

$(x - 12) (x + 3) = 0$

Simplify $(x - 12) (x + 3)$ .

Expand $(x - 12) (x + 3)$ using the FOIL Method.

Apply the distributive property.

$x (x + 3) - 12 (x + 3) = 0$

Apply the distributive property.

$x \cdot x + x \cdot 3 - 12 (x + 3) = 0$

Apply the distributive property.

$x \cdot x + x \cdot 3 - 12 x - 12 \cdot 3 = 0$

Simplify and combine like terms.

Simplify each term.

Multiply $x$ by $x$ .

$x^{2} + x \cdot 3 - 12 x - 12 \cdot 3 = 0$

Move $3$ to the left of $x$ .

$x^{2} + 3 \cdot x - 12 x - 12 \cdot 3 = 0$

Multiply $- 12$ by $3$ .

$x^{2} + 3 x - 12 x - 36 = 0$

Subtract $12 x$ from $3 x$ .

$x^{2} - 9 x - 36 = 0$

Factor $x^{2} - 9 x - 36$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 36$ and whose sum is $- 9$ .

$- 12, 3$

Write the factored form using these integers.

$(x - 12) (x + 3) = 0$

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x - 12 = 0$

$x + 3 = 0$

Set $x - 12$ equal to $0$ and solve for $x$ .

Set $x - 12$ equal to $0$ .

$x - 12 = 0$

Add $12$ to both sides of the equation.

$x = 12$

Set $x + 3$ equal to $0$ and solve for $x$ .

Set $x + 3$ equal to $0$ .

$x + 3 = 0$

Subtract $3$ from both sides of the equation.

$x = - 3$

The final solution is all the values that make $(x - 12) (x + 3) = 0$ true.

$x = 12, - 3$

Exclude the solutions that do not make $\log_{3} (x + 6) + \log_{3} (x - 6) - \log_{3} (x) = 2$ true.

$x = 12$

Solve for x log base 3 of x+6+ log base 3 of x-6- log base 3 of x=2

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