To solve this problem, we can use the exponential growth formula:

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

Where:

• $P(t)$ is the population at time $t$.
• $P_0$ is the initial population.
• $r$ is the growth rate.
• $t$ is the time elapsed.

Step 1: Identify the known values.

• $P_0 = 6900$ (population in 2005).
• $P(7) = 13500$ (population in 2012, which is 7 years after 2005).
• $t = 7$ (time in years from 2005 to 2012).

Step 2: Set up the equation for 2012 and solve for the growth rate $r$.

$13500 = 6900 \cdot e^{7r}$

Divide both sides by 6900:

$\frac{13500}{6900} = e^{7r}$

Simplify:

$\frac{13500}{6900} = \frac{135}{69} \approx 1.9565$

Now, take the natural logarithm (ln) of both sides:

$\ln(1.9565) = 7r$

$r \approx \frac{\ln(1.9565)}{7}$

Step 3: Calculate $r$.

$r \approx \frac{0.6728}{7} \approx 0.0961$

Step 4: Predict the population in 2015 (10 years after 2005).

Now, calculate $P(10)$ using the exponential growth formula:

$P(10) = 6900 \cdot e^{0.0961 \cdot 10}$

$P(10) = 6900 \cdot e^{0.961}$

Step 5: Compute $P(10)$.

$P(10) \approx 6900 \cdot 2.6146 \approx 18040$

Final Answer:

The population of rabbits in the year 2015 is approximately 18,040.

Would you like more details, or do you have any questions?

Here are 5 related questions to further your understanding:

1. How would the population change if the growth rate was higher?
2. What would the population be in 2020 using the same growth rate?
3. How can exponential decay be modeled similarly?
4. What happens to the population if the growth rate is negative?
5. How would the model change if there was a carrying capacity on the island?

Tip: Always check the units and context when interpreting the growth rate in exponential models.