We are asked to solve the equation $\ln(x+4) + \ln(x) = \ln(x+18)$ for $x$.

1. Problem Description

We are asked to solve the equation $\ln(x+4) + \ln(x) = \ln(x+18)$ for $x$.

2. Solution Steps

We are given the equation $\ln(x+4) + \ln(x) = \ln(x+18)$.

Using the property of logarithms that $\ln(a) + \ln(b) = \ln(ab)$, we can rewrite the left side of the equation as follows:

$\ln((x+4)x) = \ln(x+18)$

Since the logarithms are equal, the arguments must be equal. Therefore, we have:

$(x+4)x = x+18$

Expanding the left side, we get:

$x^2 + 4x = x+18$

Subtracting $x$ and 18 from both sides, we have:

$x^2 + 3x - 18 = 0$

Now we need to factor the quadratic equation:

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

So the possible solutions are $x = -6$ and $x = 3$.

However, we need to check if these solutions are valid in the original equation.

If $x = -6$, then $\ln(x+4) = \ln(-6+4) = \ln(-2)$, and $\ln(x) = \ln(-6)$. Since we cannot take the logarithm of a negative number, $x = -6$ is not a valid solution.

If $x = 3$, then $\ln(x+4) = \ln(3+4) = \ln(7)$, $\ln(x) = \ln(3)$, and $\ln(x+18) = \ln(3+18) = \ln(21)$.

Since $\ln(7) + \ln(3) = \ln(7 \cdot 3) = \ln(21)$, $x=3$ is a valid solution.

3. Final Answer

$x=3$