To solve the equation $0.52 = e^{-0.03x}$, follow these steps:

Step 1: Take the natural logarithm (ln) of both sides

$\ln(0.52) = \ln(e^{-0.03x})$ Using the logarithmic property $\ln(e^y) = y$, we simplify the right-hand side: $\ln(0.52) = -0.03x$

Step 2: Solve for $x$

$x = \frac{\ln(0.52)}{-0.03}$

Step 3: Calculate the logarithm and divide

$\ln(0.52) \approx -0.65393$ $x = \frac{-0.65393}{-0.03} \approx 21.80$

Thus, $x \approx 21.80$.

Would you like more details or have any questions?

Here are five related questions:

1. How do you solve exponential equations involving other bases?
2. What is the natural logarithm, and why is it important in solving such equations?
3. Can exponential functions model real-world scenarios? If yes, how?
4. How does the base $e$ influence the behavior of exponential functions?
5. What is the inverse function of an exponential function?

Tip: When solving exponential equations, always consider logarithms as the inverse of exponentials!