To solve the equation \( 0.1 = e^x \), we need to isolate \( x \).

1. Take the natural logarithm (ln) on both sides of the equation:
   \[
   \ln(0.1) = \ln(e^x)
   \]

2. Simplifying the right-hand side using the property \( \ln(e^x) = x \):
   \[
   \ln(0.1) = x
   \]

3. Now, calculate \( \ln(0.1) \):
   \[
   x = \ln(0.1) \approx -2.3026
   \]

Thus, \( x \approx -2.3026 \).

Would you like more details or have any questions?

Here are 5 related questions to expand on this topic:
1. How do we apply logarithms to solve other exponential equations?
2. What are the properties of natural logarithms (ln)?
3. How do we solve exponential equations with different bases, like \( a^x = b \)?
4. Can we solve \( 0.1 = e^{2x} \) in a similar way? 
5. How can logarithms be used in real-world exponential growth problems?

**Tip:** Natural logarithms are extremely useful when dealing with base \( e \) exponential equations, as they simplify calculations involving \( e^x \).

Learn how to solve the exponential equation 0.1 = e^x by applying natural logarithms. This solution includes step-by-step instructions on isolating x and calculating logarithms. The problem demonstrates key concepts such as logarithmic identities and exponential equations, suitable for students in Grades 9-12.

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