We are given the equation:

\[
0.97^{-4n} = 2e^{0.08n}
\]

To solve for \( n \), follow these steps:

### Step 1: Take the natural logarithm (ln) on both sides
Apply the natural logarithm to both sides of the equation to deal with the exponents:

\[
\ln(0.97^{-4n}) = \ln(2e^{0.08n})
\]

### Step 2: Use logarithmic properties
Use the logarithmic property \( \ln(a^b) = b\ln(a) \) on both sides:

\[
-4n \ln(0.97) = \ln(2) + \ln(e^{0.08n})
\]

Since \( \ln(e^x) = x \), this simplifies to:

\[
-4n \ln(0.97) = \ln(2) + 0.08n
\]

### Step 3: Solve for \( n \)
Now, isolate \( n \). First, move all terms involving \( n \) to one side:

\[
-4n \ln(0.97) - 0.08n = \ln(2)
\]

Factor out \( n \) on the left-hand side:

\[
n(-4 \ln(0.97) - 0.08) = \ln(2)
\]

Finally, solve for \( n \):

\[
n = \frac{\ln(2)}{-4 \ln(0.97) - 0.08}
\]

### Step 4: Substitute values
Now substitute the values of \( \ln(2) \) and \( \ln(0.97) \):

\[
\ln(2) \approx 0.6931, \quad \ln(0.97) \approx -0.0305
\]

Substitute these into the equation:

\[
n = \frac{0.6931}{-4(-0.0305) - 0.08}
\]

Simplify the denominator:

\[
n = \frac{0.6931}{0.122 + 0.08} = \frac{0.6931}{0.202}
\]

Finally, calculate the value of \( n \):

\[
n \approx 3.43
\]

Thus, the solution is \( n \approx 3.43 \).

---

Would you like more details or have any questions?

### Related Questions:
1. How do logarithmic properties help solve equations involving exponents?
2. What is the importance of the natural logarithm in solving exponential equations?
3. Can you show how to solve similar equations where the base is different from \(e\)?
4. What are other methods to solve exponential equations if logarithms cannot be used directly?
5. How would the solution change if the base of the exponential function was \(10\) instead of \(e\)?

**Tip:** Always remember to check your final solution by substituting it back into the original equation!

Learn how to solve the exponential equation 0.97^(-4n) = 2e^(0.08n) using logarithmic properties and algebra. This step-by-step solution involves natural logarithms, exponent rules, and solving for n. Perfect for high school students in Grades 10-12.

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