The problem asks us to solve the equation $\ln(x+2) + \ln(x) = \ln(x+30)$ for $x$.

1. Problem Description

The problem asks us to solve the equation $\ln(x+2) + \ln(x) = \ln(x+30)$ for $x$.

2. Solution Steps

We are given the equation:

$\ln(x+2) + \ln(x) = \ln(x+30)$

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

$\ln(x(x+2)) = \ln(x+30)$

Since the logarithms are equal, we can equate the arguments:

$x(x+2) = x+30$

Expanding the left side, we get:

$x^2 + 2x = x + 30$

Now, move all terms to one side to form a quadratic equation:

$x^2 + 2x - x - 30 = 0$

$x^2 + x - 30 = 0$

We can factor this quadratic equation:

$(x+6)(x-5) = 0$

This gives us two possible solutions for $x$:

$x+6 = 0 \implies x = -6$

$x-5 = 0 \implies x = 5$

Now, we need to check if these solutions are valid by plugging them back into the original equation. Since the argument of a logarithm must be positive, we need $x+2 > 0$, $x > 0$, and $x+30 > 0$. This means we need $x > -2$ and $x > 0$. Thus, $x$ must be greater than $0$.

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

If $x = 5$, then $\ln(5+2) = \ln(7)$, $\ln(5)$, and $\ln(5+30) = \ln(35)$ are all defined. Thus, $x = 5$ is a valid solution.

Therefore, the solution is $x=5$.

3. Final Answer

$x = 5$