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

1. Problem Description

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

2. Solution Steps

We use the property of logarithms that $\ln(a) + \ln(b) = \ln(ab)$ to combine the left side of the equation. This gives us

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

Since the logarithms are equal, their arguments must be equal. Therefore,

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

Expanding the left side, we have

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

Subtracting $x+18$ from both sides, we get a quadratic equation:

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

We can factor this quadratic as follows:

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

The solutions to this equation are $x=-6$ and $x=3$.

However, we must check if these solutions are valid in the original equation. We cannot take the logarithm of a negative number or zero.

If $x=-6$, then $\ln(x) = \ln(-6)$ and $\ln(x+4) = \ln(-2)$, which are both undefined. Therefore, $x=-6$ is not a valid solution.

If $x=3$, then $\ln(x) = \ln(3)$, $\ln(x+4) = \ln(7)$, and $\ln(x+18) = \ln(21)$. Since $3, 7, 21$ are all positive, $x=3$ is a valid solution.

We check our answer by substituting $x=3$ into the original equation:

$\ln(3+4) + \ln(3) = \ln(7) + \ln(3) = \ln(7 \cdot 3) = \ln(21)$.

$\ln(3+18) = \ln(21)$.

Since both sides are equal, $x=3$ is the solution.

3. Final Answer

$x=3$