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!

Learn how to solve the exponential equation 0.52 = e^-0.03x using natural logarithms. This detailed step-by-step solution explains logarithmic properties, exponential function properties, and how to isolate variables. Suitable for students in Grades 10-12.

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