To determine how long it will take for the infected population to grow from 3 to 800 people in an epidemic with an exponential growth rate, we can use the exponential growth formula:

$P(t) = P_0 e^{kt}$

where:

• $P(t)$ is the population at time $t$,
• $P_0$ is the initial population,
• $k$ is the growth rate,
• $t$ is the time.

Given:

• $P_0 = 3$,
• $P(t) = 800$,
• $k = 0.32$ per annum.

We need to find $t$. Start by plugging the given values into the formula:

$800 = 3 e^{0.32t}$

Next, solve for $t$:

1. Divide both sides by 3: $\frac{800}{3} = e^{0.32t}$

2. Simplify the fraction: $266.67 = e^{0.32t}$

3. Take the natural logarithm of both sides to solve for $t$: $\ln(266.67) = \ln(e^{0.32t})$

4. Use the property of logarithms that $\ln(e^x) = x$: $\ln(266.67) = 0.32t$

5. Solve for $t$: $t = \frac{\ln(266.67)}{0.32}$

6. Calculate the natural logarithm and division: $t \approx \frac{5.586}{0.32} \approx 17.46$

Therefore, it will take approximately 17.46 years for the infected population to grow from 3 to 800 people.

Would you like more details or have any questions?

Here are some related questions:

1. What is exponential growth and how is it different from linear growth?
2. How does the growth rate (k) affect the time it takes for the population to reach a certain size?
3. How can we determine the doubling time for an epidemic given the growth rate?
4. What other factors might influence the spread of an epidemic in a real-world scenario?
5. How can public health interventions impact the exponential growth rate of an epidemic?

Tip: Always check units when calculating exponential growth problems to ensure they are consistent (e.g., time units per annum).