The problem indicates that the deer population grows exponentially, and the function modeling this population can be written as:

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

Where:

• $P(t)$ is the population at time $t$,
• $P_0$ is the initial population (at $t = 0$),
• $r$ is the growth rate,
• $t$ is time in years.

From the problem, we know:

• $P_0 = 10,000$ (initial population at $t = 0$),
• $P(4) = 26,000$ (population after 4 years, at $t = 4$).

We need to find the growth rate $r$.

Step 1: Set up the equation using the given values

From the exponential growth formula:

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

Substitute the values for $P(4) = 26,000$, $P_0 = 10,000$, and $t = 4$:

$26,000 = 10,000 e^{4r}$

Step 2: Solve for $r$

First, divide both sides of the equation by 10,000 to isolate the exponential term:

$\frac{26,000}{10,000} = e^{4r}$ $2.6 = e^{4r}$

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

$\ln(2.6) = 4r$ $r = \frac{\ln(2.6)}{4}$

Step 3: Calculate the value of $r$

Using a calculator:

$\ln(2.6) \approx 0.9555$ $r = \frac{0.9555}{4} \approx 0.2389$

Rounding to the nearest whole number percentage:

$r \times 100 \approx 24\%$

Final Answer: The growth rate is approximately 24% per year.

Would you like further details or have any questions?

Here are 5 related questions:

1. How does the natural logarithm help in solving exponential equations?
2. What would happen to the growth rate if the population doubled in 4 years instead of increasing to 26,000?
3. How would you graph an exponential growth function for population modeling?
4. What if the population had been declining? How would that affect the equation?
5. Can you explain how exponential growth differs from linear growth?

Tip: Exponential growth means that as time passes, the rate of increase becomes faster, unlike linear growth, which remains constant over time.