Let's solve the given inequalities:

(a) $\ln(x) < 0$

To solve this inequality, we need to determine for what values of $x$, the natural logarithm of $x$ is less than zero.

1. Recall that the natural logarithm function $\ln(x)$ is less than zero when $0 < x < 1$, because the graph of $\ln(x)$ crosses the x-axis at $x = 1$, and for $x > 1$, $\ln(x)$ is positive.

Thus, the solution to the inequality $\ln(x) < 0$ is: $(0, 1)$ in interval notation.

(b) $e^x > 3$

To solve this inequality, follow these steps:

1. Take the natural logarithm of both sides of the inequality: $\ln(e^x) > \ln(3)$
2. Simplifying the left-hand side using the property $\ln(e^x) = x$, we get: $x > \ln(3)$
3. Since $\ln(3)$ is a constant (approximately 1.0986), the solution is: $x > \ln(3)$ In interval notation, this is written as: $(\ln(3), \infty)$

Let me know if you'd like further details or explanations! Here are some related questions you can explore:

1. How do you graph the function $\ln(x)$?
2. Can $\ln(x)$ ever be negative for values greater than 1?
3. What happens to $e^x$ as $x \to -\infty$?
4. How would you solve $e^x \geq 3$ instead of $e^x > 3$?
5. What is the general method for solving exponential inequalities?

Tip: For exponential inequalities like $e^x > a$, you can always take the natural logarithm of both sides to simplify the inequality.