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 $e^x$.