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.