To solve the equation $e^x - 9 = 0$:

1. Add 9 to both sides of the equation:

$e^x = 9$

2. Now, take the natural logarithm (ln) of both sides to solve for $x$:

$\ln(e^x) = \ln(9)$

3. Since $\ln(e^x) = x$, this simplifies to:

$x = \ln(9)$

4. So, the solution is:

$x = \ln(9) \approx 2.1972$

Let me know if you'd like more details or have any questions!

Here are 5 related questions for further exploration:

1. How do you solve exponential equations in general?
2. What is the derivative of $e^x$?
3. What are the properties of natural logarithms?
4. How does $\ln(e^x)$ simplify to $x$?
5. How can you apply logarithms to different bases in equations?

Tip: The natural logarithm $\ln(x)$ is the inverse function of $e^x$, and it's useful for solving equations where the variable is in the exponent.