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The problem provided involves an exponential decay model for a frog population at Enormous State University (ESU). The questions are as follows:

1. **Problem (a): Find an exponential model for the frog population, with \( t \) in years.**

   We know:
   - Two years ago (\( t = 0 \)), 40,000 frogs were tagged.
   - This year (\( t = 2 \)), 27,000 frogs were tagged.

   The exponential decay model generally takes the form:
   \[
   f(t) = f_0 \cdot e^{-kt}
   \]
   where \( f_0 \) is the initial population, \( k \) is the decay constant, and \( t \) is time in years.

   To find \( k \), we use the data points: 
   - \( f(0) = 40,000 \)
   - \( f(2) = 27,000 \)

   Substituting into the model for \( t = 2 \):
   \[
   27,000 = 40,000 \cdot e^{-2k}
   \]

   Solving for \( k \):
   \[
   \frac{27,000}{40,000} = e^{-2k}
   \]
   \[
   \frac{27}{40} = e^{-2k}
   \]
   Taking the natural logarithm on both sides:
   \[
   \ln\left(\frac{27}{40}\right) = -2k
   \]
   \[
   k = -\frac{1}{2} \ln\left(\frac{27}{40}\right)
   \]
   Approximate \( k \) using a calculator:
   \[
   k \approx 0.185
   \]
   
   Therefore, the exponential model is approximately:
   \[
   f(t) = 40,000 \cdot e^{-0.185t}
   \]

---

2. **Problem (b): Assuming exponential population decrease and that this year’s tags fall off, how many tags should Epsilon Delta order for next year’s pledge class?**

   Next year corresponds to \( t = 3 \). Using the model \( f(t) = 40,000 \cdot e^{-0.185t} \):
   \[
   f(3) = 40,000 \cdot e^{-0.185 \cdot 3}
   \]
   Approximate this:
   \[
   f(3) \approx 40,000 \cdot e^{-0.555} \approx 40,000 \cdot 0.573 \approx 22,920
   \]

   So, they should order approximately **22,920 tags**.

---

3. **Problem (c): If the current trend continues, what is the first year in which fewer than 1,000 tags are required?**

   We need to find \( t \) when \( f(t) < 1,000 \). From the model:
   \[
   1,000 = 40,000 \cdot e^{-0.185t}
   \]
   Solving for \( t \):
   \[
   \frac{1,000}{40,000} = e^{-0.185t}
   \]
   \[
   \frac{1}{40} = e^{-0.185t}
   \]
   Taking the natural logarithm on both sides:
   \[
   \ln\left(\frac{1}{40}\right) = -0.185t
   \]
   \[
   t = \frac{\ln(1/40)}{-0.185} \approx \frac{-3.689}{-0.185} \approx 19.94
   \]

   So, the first year when fewer than 1,000 tags are required is around **year 20**.

---

Would you like more details or have any questions?

Here are 5 related questions to expand on this topic:
1. How does changing the decay rate \( k \) affect the population model?
2. What happens if the initial population is increased to 50,000? How does the model change?
3. How can we model population growth instead of decay?
4. What would the population be in year 10 using this exponential model?
5. How would you calculate the half-life of the frog population in this scenario?

**Tip:** Always check units carefully in exponential decay problems, as the decay constant \( k \) can have different interpretations depending on the time units used.

Frogs have been dying like flies at the Enormous State University (ESU) campus. Each year, the pledge class of the Epsilon Delta fraternity is instructed to tag all the frogs residing on the ESU campus. Two years ago (t=0), they managed to tag 40,000 frogs. This year (t=2), they tagged 27,000 frogs. (a) Find an exponential model for the frog population, with t in years. (b) Assuming exponential population decrease and that all this year's tags fall off, how many tags should Epsilon Delta order for next year’s pledge class? (c) If the current trend continues, what is the first year in which fewer than 1,000 tags are required?

Explore how to model the frog population decline at Enormous State University using an exponential decay model. This problem involves calculating the exponential decay constant and predicting future frog populations. Suitable for high school students in Grades 10-12.

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