To solve this problem, we will model the population of bats using the exponential growth formula:

$P(t) = P_0 \cdot e^{kt}$

Where:

• $P(t)$ is the population at time $t$,
• $P_0$ is the initial population,
• $k$ is the growth rate, and
• $t$ is the time in years.

We are given:

• After 1 year, $P(1) = 70$,
• After 4 years, $P(4) = 560$.

Step 1: Set up two equations based on the data.

For $t = 1$: $P(1) = P_0 \cdot e^{k \cdot 1} = 70 \quad \Rightarrow \quad P_0 \cdot e^k = 70$

For $t = 4$: $P(4) = P_0 \cdot e^{k \cdot 4} = 560 \quad \Rightarrow \quad P_0 \cdot e^{4k} = 560$

Step 2: Solve the system of equations.

To eliminate $P_0$, divide the second equation by the first: $\frac{P_0 \cdot e^{4k}}{P_0 \cdot e^k} = \frac{560}{70}$

This simplifies to: $e^{3k} = 8$

Take the natural logarithm (ln) of both sides: $3k = \ln(8)$ $k = \frac{\ln(8)}{3} \approx \frac{2.0794}{3} \approx 0.6931$

Step 3: Find $P_0$.

Using the first equation $P_0 \cdot e^k = 70$, substitute $k \approx 0.6931$: $P_0 \cdot e^{0.6931} = 70 \quad \Rightarrow \quad P_0 \cdot 2 = 70 \quad \Rightarrow \quad P_0 = \frac{70}{2} = 35$

Step 4: Calculate the population after 7 years.

Now that we know $P_0 = 35$ and $k \approx 0.6931$, we can find the population after 7 years using the formula $P(7) = P_0 \cdot e^{k \cdot 7}$:

$P(7) = 35 \cdot e^{0.6931 \cdot 7} = 35 \cdot e^{4.8517} \approx 35 \cdot 127.017 \approx 4445.6$

So, the population after 7 years is approximately 4446 bats.

Would you like further clarification or additional details?

Related Questions:

1. How does exponential growth differ from linear growth?
2. Can exponential growth continue indefinitely in real-world scenarios?
3. How does the growth rate affect the speed of population increase?
4. How would the model change if we had a carrying capacity in the environment?
5. What is the doubling time for this colony of bats?

Tip:

Always double-check calculations when dealing with exponentials, especially when approximating values.