Solve for x log of 3x+ log of x-4 = log of 15

$\log (3 x) + \log (x - 4) = \log (15)$

Simplify the left side.

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

$\log (3 x (x - 4)) = \log (15)$

Apply the distributive property.

$\log (3 x \cdot x + 3 x \cdot - 4) = \log (15)$

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

Move $x$ .

$\log (3 (x \cdot x) + 3 x \cdot - 4) = \log (15)$

Multiply $x$ by $x$ .

$\log (3 x^{2} + 3 x \cdot - 4) = \log (15)$

Multiply $- 4$ by $3$ .

$\log (3 x^{2} - 12 x) = \log (15)$

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

$3 x^{2} - 12 x = 15$

Solve for $x$ .

Subtract $15$ from both sides of the equation.

$3 x^{2} - 12 x - 15 = 0$

Factor the left side of the equation.

Factor $3$ out of $3 x^{2} - 12 x - 15$ .

Factor $3$ out of $3 x^{2}$ .

$3 (x^{2}) - 12 x - 15 = 0$

Factor $3$ out of $- 12 x$ .

$3 (x^{2}) + 3 (- 4 x) - 15 = 0$

Factor $3$ out of $- 15$ .

$3 x^{2} + 3 (- 4 x) + 3 \cdot - 5 = 0$

Factor $3$ out of $3 x^{2} + 3 (- 4 x)$ .

$3 (x^{2} - 4 x) + 3 \cdot - 5 = 0$

Factor $3$ out of $3 (x^{2} - 4 x) + 3 \cdot - 5$ .

$3 (x^{2} - 4 x - 5) = 0$

Factor.

Factor $x^{2} - 4 x - 5$ 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 $- 5$ and whose sum is $- 4$ .

$- 5, 1$

Write the factored form using these integers.

$3 ((x - 5) (x + 1)) = 0$

Remove unnecessary parentheses.

$3 (x - 5) (x + 1) = 0$

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

$x - 5 = 0$

$x + 1 = 0$

Set $x - 5$ equal to $0$ and solve for $x$ .

Set $x - 5$ equal to $0$ .

$x - 5 = 0$

Add $5$ to both sides of the equation.

$x = 5$

Set $x + 1$ equal to $0$ and solve for $x$ .

Set $x + 1$ equal to $0$ .

$x + 1 = 0$

Subtract $1$ from both sides of the equation.

$x = - 1$

The final solution is all the values that make $3 (x - 5) (x + 1) = 0$ true.

$x = 5, - 1$

Exclude the solutions that do not make $\log (3 x) + \log (x - 4) = \log (15)$ true.

$x = 5$

Solve for x log of 3x+ log of x-4 = log of 15

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