Solve for x log of x+ log of x+5 = log of 6x+12

$\log (x) + \log (x + 5) = \log (6 x + 12)$

Simplify the left side.

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

$\log (x (x + 5)) = \log (6 x + 12)$

Apply the distributive property.

$\log (x \cdot x + x \cdot 5) = \log (6 x + 12)$

Simplify the expression.

Multiply $x$ by $x$ .

$\log (x^{2} + x \cdot 5) = \log (6 x + 12)$

Move $5$ to the left of $x$ .

$\log (x^{2} + 5 x) = \log (6 x + 12)$

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

$x^{2} + 5 x = 6 x + 12$

Solve for $x$ .

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

Subtract $6 x$ from both sides of the equation.

$x^{2} + 5 x - 6 x = 12$

Subtract $6 x$ from $5 x$ .

$x^{2} - x = 12$

Subtract $12$ from both sides of the equation.

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

Factor $x^{2} - x - 12$ 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 $- 12$ and whose sum is $- 1$ .

$- 4, 3$

Write the factored form using these integers.

$(x - 4) (x + 3) = 0$

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

$x - 4 = 0$

$x + 3 = 0$

Set $x - 4$ equal to $0$ and solve for $x$ .

Set $x - 4$ equal to $0$ .

$x - 4 = 0$

Add $4$ to both sides of the equation.

$x = 4$

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 $(x - 4) (x + 3) = 0$ true.

$x = 4, - 3$

Exclude the solutions that do not make $\log (x) + \log (x + 5) = \log (6 x + 12)$ true.

$x = 4$

Solve for x log of x+ log of x+5 = log of 6x+12

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