The natural logarithm of \( e^{-kx} \) can be simplified as follows:

\[
\ln(e^{-kx}) = -kx
\]

This is because the logarithm of \( e^y \) is simply \( y \), and in this case, \( y = -kx \). So:

\[
\ln(e^{-kx}) = -kx
\]

Would you like further details or have any questions?

Here are 5 related questions:
1. What is the derivative of \( \ln(e^{-kx}) \) with respect to \( x \)?
2. How would you integrate \( \ln(e^{-kx}) \)?
3. What happens if you take the logarithm of \( e^{kx} \) instead of \( e^{-kx} \)?
4. How does the property \( \ln(ab) = \ln(a) + \ln(b) \) apply to more complex logarithmic functions?
5. How can you apply logarithmic identities to solve exponential equations?

**Tip:** The natural logarithm and exponential functions are inverses of each other, which is why they "cancel out" in expressions like \( \ln(e^x) = x \).

Learn how to simplify the expression ln(e^-kx) to -kx. This problem explores the relationship between logarithms and exponential functions, using the formula ln(e^y) = y. Suitable for high school students in Grades 9-12.

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