We are given the population equation $P(t) = 300e^{0.01t}$, where $P(t)$ is the population at time $t$ in days, and we want to find the time $t$ when the population reaches 1500.

We set $P(t) = 1500$ and solve for $t$.

$1500 = 300e^{0.01t}$

Step 1: Divide both sides by 300

$\frac{1500}{300} = e^{0.01t}$ $5 = e^{0.01t}$

Step 2: Take the natural logarithm (ln) of both sides

$\ln(5) = \ln(e^{0.01t})$ $\ln(5) = 0.01t$

Step 3: Solve for $t$

$t = \frac{\ln(5)}{0.01}$

Now, we can compute this:

$t = \frac{\ln(5)}{0.01} \approx \frac{1.6094}{0.01} = 160.94$

Rounding to one decimal place, $t \approx 160.9$ days.

Thus, it will take approximately 160.9 days for the population to reach 1500.

Would you like more details or further clarification?

Here are 5 questions that expand upon this concept:

1. How does the population change over time when $t$ increases?
2. What would the population be at $t = 0$ days?
3. How would you find the rate of growth of the population at any given time $t$?
4. If the growth rate constant changed to 0.02, how would that affect the time to reach 1500?
5. How long would it take for the population to double?

Tip: Always check the base of the logarithm you use to ensure it's appropriate for the problem (e.g., natural logarithm for base $e$).