To solve the equation $\ln(x^2) + \ln(x) = 0$, we can first simplify it using logarithmic properties.

Step-by-step solution:

1. Combine the logarithmic terms: Recall that $\ln(a) + \ln(b) = \ln(ab)$. Thus, $\ln(x^2) + \ln(x) = \ln(x^2 \cdot x) = \ln(x^3)$ The equation now becomes: $\ln(x^3) = 0$

2. Exponentiate both sides: To remove the logarithm, exponentiate both sides of the equation: $e^{\ln(x^3)} = e^0$ This simplifies to: $x^3 = 1$

3. Solve for $x$: To solve for $x$, take the cube root of both sides: $x = 1$

Thus, the solution to the equation is $x = 1$.

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Do you want more details on any of the steps or further clarification?

Here are some related questions:

1. What is the derivative of $\ln(x^2) + \ln(x)$?
2. How does the property $\ln(a) + \ln(b) = \ln(ab)$ work?
3. What happens if $\ln(x^2) + \ln(x) = \ln(c)$ instead of 0?
4. Can you solve $\ln(x^n) + \ln(x) = 0$ for any value of $n$?
5. What is the general rule for solving logarithmic equations?

Tip: Remember, $\ln(x)$ is only defined for $x > 0$, so when solving logarithmic equations, check that your solutions are in the domain.