The given equation is:

$\ln(x) + \ln(x - 4) = \ln(6x)$

Step-by-step solution:

1. Use the logarithmic property that $\ln(a) + \ln(b) = \ln(ab)$:

   $\ln(x) + \ln(x - 4) = \ln(x(x - 4)) = \ln(x^2 - 4x)$

   So the equation becomes:

   $\ln(x^2 - 4x) = \ln(6x)$

2. Since the logarithms are equal, the arguments must also be equal:

   $x^2 - 4x = 6x$

3. Simplify the quadratic equation:

   $x^2 - 4x - 6x = 0$

   $x^2 - 10x = 0$

4. Factor the equation:

   $x(x - 10) = 0$

5. Solve for $x$:

   $x = 0 \quad \text{or} \quad x = 10$

6. Check for extraneous solutions:

   • $x = 0$ is not a valid solution because $\ln(0)$ is undefined.
   • Therefore, the only valid solution is $x = 10$.

Thus, $x = 10$.

Would you like more details or have any questions? Here are five related questions to expand your understanding:

1. How does the property $\ln(a) + \ln(b) = \ln(ab)$ apply to logarithmic equations?
2. What are the general steps for solving equations involving logarithms?
3. Can logarithmic functions have negative or zero inputs?
4. What is an extraneous solution, and why did we discard $x = 0$?
5. How do we convert logarithmic equations to exponential form?

Tip: Always check for extraneous solutions when solving logarithmic equations because logarithms are undefined for non-positive numbers.