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.

Learn how to solve the exponential equation e^x - 9 = 0 using natural logarithms. This step-by-step guide covers key concepts such as exponential functions, logarithms, and their properties. Suitable for high school students in Grades 10-12.

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