To express the equation for $P$ (future population) in the exponential decay model given the information provided, we start with the general exponential decay formula:

$P = P_0 \cdot e^{kt}$

Given the following dataTo express the equation for $P$ (future population) in the exponential decay model given the information provided, we start with the general exponential decay formula:

$P = P_0 \cdot e^{kt}$

Given the following

• Initial population $P_0 = 3000$
• Annual decay rate $r = -3.8\% = -0.038$
• Time in years $t$

We substitute these values into the formula. The decay rate $k$ is $-0.038$:

$P = 3000 \cdot e^{-0.038t}$

So the exponential decay model for the future population $P$ is:

$P = 3000 \cdot e^{-0.038t}$

To find the population in 4 years:

1. Substitute $t = 4$:

$P(4) = 3000 \cdot e^{-0.038 \cdot 4}$

2. Simplify the exponent:

$P(4) = 3000 \cdot e^{-0.152}$

3. Calculate $e^{-0.152} \approx 0.859$:

$P(4) = 3000 \cdot 0.859 \approx 2577$

Therefore, the population in 4 years will be approximately 2577 students (rounded to the nearest student).

The equation you need is:

$P = 3000 \cdot e^{-0.038t}$

If you have any questions or need more details, let me know!

Here are 5 more questions you might consider:

1. How do you convert a percentage decay rate to a decay constant?
2. What is the difference between exponential growth and exponential decay?
3. How do you use logarithms to solve for time in an exponential equation?
4. Can you give an example of exponential growth in a different context?
5. How does changing the initial population affect the future population in exponential models?

Tip: When working with exponential models, it's important to remember that the base of the natural logarithm $e$ (approximately 2.718) is a constant used in these calculations.