Solve for x log of x+ log of x+4 = log of 12

$\log (x) + \log (x + 4) = \log (12)$

Simplify the left side.

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

$\log (x (x + 4)) = \log (12)$

Apply the distributive property.

$\log (x \cdot x + x \cdot 4) = \log (12)$

Simplify the expression.

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Multiply $x$ by $x$ .

$\log (x^{2} + x \cdot 4) = \log (12)$

Move $4$ to the left of $x$ .

$\log (x^{2} + 4 x) = \log (12)$

For the equation to be equal, the argument of the logarithms on both sides of the equation must be equal.

$x^{2} + 4 x = 12$

Solve for $x$ .

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Move $12$ to the left side of the equation by subtracting it from both sides.

$x^{2} + 4 x - 12 = 0$

Factor $x^{2} + 4 x - 12$ 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 $- 12$ and whose sum is $4$ .

$- 2, 6$

Write the factored form using these integers.

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

Set $x - 2$ equal to $0$ and solve for $x$ .

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Set the factor equal to $0$ .

$x - 2 = 0$

Add $2$ to both sides of the equation.

$x = 2$

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

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Set the factor equal to $0$ .

$x + 6 = 0$

Subtract $6$ from both sides of the equation.

$x = - 6$

The solution is the result of $x - 2 = 0$ and $x + 6 = 0$ .

$x = 2, - 6$

Exclude the solutions that do not make $\log (x) + \log (x + 4) = \log (12)$ true.

$x = 2$

Solve for x log of x+ log of x+4 = log of 12