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Let's solve each part of the problem step by step.

The logistic model given is:
\[ P(t) = \frac{840}{1 + 14e^{-0.15t}} \]

### Part (a)
**Find the initial number of elk planted in the forest.**

The initial number of elk corresponds to \( t = 0 \).

\[ P(0) = \frac{840}{1 + 14e^{-0.15 \cdot 0}} = \frac{840}{1 + 14} = \frac{840}{15} = 56 \]

So, the initial number of elk is **56**.

### Part (b)
**Determine the carrying capacity of the forest.**

The carrying capacity \( K \) of a logistic model is the maximum population the environment can sustain, which is given by the numerator of the logistic function.

\[ K = 840 \]

So, the carrying capacity of the forest is **840** elk.

### Part (c)
**Determine when the population reaches \( \frac{1}{2} \) of the carrying capacity.**

Half of the carrying capacity is \( \frac{840}{2} = 420 \).

Set \( P(t) = 420 \) and solve for \( t \):

\[ 420 = \frac{840}{1 + 14e^{-0.15t}} \]

Solving for \( t \):

\[ 1 + 14e^{-0.15t} = 2 \]
\[ 14e^{-0.15t} = 1 \]
\[ e^{-0.15t} = \frac{1}{14} \]
\[ -0.15t = \ln\left(\frac{1}{14}\right) \]
\[ t = \frac{\ln(14)}{0.15} \approx \frac{2.639}{0.15} \approx 17.59 \]

So, the population reaches half of the carrying capacity at approximately **17.59 years**.

### Part (d)
**Determine the number of years it takes for the population to reach 750 animals.**

Set \( P(t) = 750 \) and solve for \( t \):

\[ 750 = \frac{840}{1 + 14e^{-0.15t}} \]

Solving for \( t \):

\[ 1 + 14e^{-0.15t} = \frac{840}{750} \]
\[ 1 + 14e^{-0.15t} = \frac{28}{25} \]
\[ 14e^{-0.15t} = \frac{28}{25} - 1 \]
\[ 14e^{-0.15t} = \frac{3}{25} \]
\[ e^{-0.15t} = \frac{3}{350} \]
\[ -0.15t = \ln\left(\frac{3}{350}\right) \]
\[ t = \frac{\ln\left(\frac{350}{3}\right)}{0.15} \approx \frac{5.857}{0.15} \approx 39.05 \]

So, it takes approximately **39.05 years** for the population to reach 750 elk.

### Part (e)
**Give the meaningful domain and range of the logistic model of elk.**

**Domain**: The domain represents the time \( t \) and since \( t \) cannot be negative in this context, the domain is:

\[ t \geq 0 \]

**Range**: The range represents the population \( P(t) \). The population starts at 56 (initial number of elk) and approaches the carrying capacity of 840 as \( t \) increases. Therefore, the range is:

\[ 56 \leq P(t) \leq 840 \]

So, the meaningful domain and range are:
- Domain: \( t \geq 0 \)
- Range: \( 56 \leq P(t) \leq 840 \)

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

Here are some questions you might want to ask:
1. Can you explain the logistic model in more detail?
2. How do you solve the equation involving \( e \) in logistic models?
3. What does the carrying capacity represent in real-world terms?
4. Why is the initial population important in logistic growth?
5. How does the logistic model differ from exponential growth?

**Tip**: Always check the context of your mathematical model to ensure your solutions make sense within that context. For example, negative time values are typically not meaningful in growth models.

Explore a detailed solution to a logistic growth model problem involving elk population dynamics. Learn how to find the initial elk population, carrying capacity, and times when the population reaches specific levels. Suitable for advanced high school students studying mathematical models.

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