We are given the exponential model $A = 481.5 e^{0.01t}$, where $A$ is the population of a country in millions, and $t$ is the number of years after 2003. The goal is to determine the year when the population will be 576 million.

We can follow these steps to solve the problem:

1. Set $A = 576$ in the given model: $576 = 481.5 e^{0.01t}$

2. Solve for $t$. First, divide both sides by 481.5: $\frac{576}{481.5} = e^{0.01t}$ Simplifying: $1.196 = e^{0.01t}$

3. Take the natural logarithm (ln) of both sides: $\ln(1.196) = 0.01t$ Using a calculator for the natural logarithm of 1.196: $0.178 = 0.01t$

4. Solve for $t$: $t = \frac{0.178}{0.01} = 17.8$

Thus, $t \approx 18$ years. Since $t$ represents the number of years after 2003, we add 18 to 2003:

$2003 + 18 = 2021$

Therefore, the population of the country will be 576 million in the year 2021.

Would you like further details or explanations?

Here are 5 related questions for deeper understanding:

1. What is the significance of the constant 481.5 in the model?
2. How does the base $e$ affect the population growth over time?
3. What happens to the population if the growth rate (0.01) changes to a higher value?
4. How can you adjust the formula if you are calculating years from a different starting point?
5. How would you use this model to predict the population in future years beyond 576 million?

Tip: When solving exponential growth models, always remember to take the natural logarithm to handle equations involving the exponential function $e^x$.