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

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

Simplify the left side.

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

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

Apply the distributive property.

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

Multiply $x$ by $x$ by adding the exponents.

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Move $x$ .

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

Multiply $x$ by $x$ .

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

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

$6^{2} = 2 x^{2} - 6 x$

Solve for $x$

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Raise $6$ to the power of $2$ .

$36 = 2 x^{2} - 6 x$

Rewrite the equation as $2 x^{2} - 6 x = 36$ .

$2 x^{2} - 6 x = 36$

Move $36$ to the left side of the equation by subtracting it from both sides.

$2 x^{2} - 6 x - 36 = 0$

Factor the left side of the equation.

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Factor $2$ out of $2 x^{2} - 6 x - 36$ .

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

$2 (x^{2}) - 6 x - 36 = 0$

Factor $2$ out of $- 6 x$ .

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

Factor $2$ out of $- 36$ .

$2 x^{2} + 2 (- 3 x) + 2 \cdot - 18 = 0$

Factor $2$ out of $2 x^{2} + 2 (- 3 x)$ .

$2 (x^{2} - 3 x) + 2 \cdot - 18 = 0$

Factor $2$ out of $2 (x^{2} - 3 x) + 2 \cdot - 18$ .

$2 (x^{2} - 3 x - 18) = 0$

Factor.

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Factor $x^{2} - 3 x - 18$ using the AC method.

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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 $- 18$ and whose sum is $- 3$ .

$- 6, 3$

Write the factored form using these integers.

$2 ((x - 6) (x + 3)) = 0$

Remove unnecessary parentheses.

$2 (x - 6) (x + 3) = 0$

Divide both sides of the equation by $2$ . Dividing $0$ by any non-zero number is $0$ .

$(x - 6) (x + 3) = 0$

Set $x - 6$ equal to $0$ and solve for $x$ .

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Set the factor equal to $0$ .

$x - 6 = 0$

Add $6$ to both sides of the equation.

$x = 6$

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

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Set the factor equal to $0$ .

$x + 3 = 0$

Subtract $3$ from both sides of the equation.

$x = - 3$

The solution is the result of $x - 6 = 0$ and $x + 3 = 0$ .

$x = 6, - 3$

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

$x = 6$

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

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