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The given population model is:

\[
P(t) = \frac{20,000}{1 + 4e^{-0.5t}}
\]

where \(P(t)\) represents the population at time \(t\), and \(t\) is measured in years.

### a) **Current Population (t = 0)**

To find the population at \(t = 0\), substitute \(t = 0\) into the equation:

\[
P(0) = \frac{20,000}{1 + 4e^{-0.5(0)}}
\]

Since \(e^0 = 1\), this simplifies to:

\[
P(0) = \frac{20,000}{1 + 4(1)} = \frac{20,000}{1 + 4} = \frac{20,000}{5} = 4,000
\]

So, the current population is **4,000**.

---

### b) **Population in 2 Years (t = 2)**

To find the population at \(t = 2\), substitute \(t = 2\) into the equation:

\[
P(2) = \frac{20,000}{1 + 4e^{-0.5(2)}}
\]

First, calculate the exponent:

\[
e^{-1} \approx 0.3679
\]

Now substitute that value:

\[
P(2) = \frac{20,000}{1 + 4(0.3679)} = \frac{20,000}{1 + 1.4716} = \frac{20,000}{2.4716} \approx 8,093
\]

So, the population in 2 years is approximately **8,093**.

---

Would you like more details on the solution or further clarifications?

### Related Questions:
1. What happens to the population as \(t \to \infty\)?
2. How would the equation change if the growth rate doubled?
3. What would the population be after 5 years?
4. How would you interpret the constant "4" in the model?
5. How does changing the initial population affect the model?

### Tip:
When solving exponential models, always simplify the exponent first before applying the base \(e\) to avoid calculation errors.

A population is modeled by P(t) = 20,000 / (1 + 4e^(-0.5t)). What is the current (t = 0) population? What is the population in 2 years?

Discover how to find the current and future populations using the logistic growth model P(t) = 20,000 / (1 + 4e^(-0.5t)). This example involves calculating exponential decay for t = 0 and t = 2. Ideal for students in Grades 11-12 studying calculus or exponential growth models.

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