To find the exponential growth model for the given data, we'll use the continuous exponential model $f(t) = Ie^{pt}$, where:

• $I$ is the initial number of cases
• $p$ is the growth rate
• $t$ is the time in days
• $f(t)$ is the number of cases at time $t$

We are given:

• Initially ($t = 0$), there are 21 confirmed cases, so $I = 21$.
• After 7 days ($t = 7$), the number of confirmed cases is 45.

We need to find the growth rate $p$.

First, let's set up the equation with the given values:

$45 = 21e^{7p}$

Next, solve for $p$:

1. Divide both sides by 21 to isolate the exponential term:

$\frac{45}{21} = e^{7p}$

2. Simplify the fraction:

$\frac{45}{21} = \frac{15}{7}$

So,

$\frac{15}{7} = e^{7p}$

3. Take the natural logarithm of both sides to solve for $p$:

$\ln\left(\frac{15}{7}\right) = \ln(e^{7p})$

4. Simplify the right-hand side using the property of logarithms ($\ln(e^x) = x$):

$\ln\left(\frac{15}{7}\right) = 7p$

5. Solve for $p$:

$p = \frac{\ln\left(\frac{15}{7}\right)}{7}$

Let's calculate this value:

$p = \frac{\ln\left(\frac{15}{7}\right)}{7} \approx \frac{\ln(2.142857)}{7} \approx \frac{0.76214}{7} \approx 0.10888$

Thus, the exponential growth rate $p$ is approximately 0.10888.

Therefore, the exponential growth model for the number of confirmed cases is:

$f(t) = 21e^{0.10888t}$

Would you like more details or have any questions?

Here are some questions you might consider next:

1. How does the exponential growth model compare to a linear growth model?
2. How would you apply this model to predict future cases?
3. What are some limitations of using an exponential growth model for disease outbreaks?
4. How can this model be adjusted for a different initial number of cases?
5. What does the growth rate $p$ represent in real-world terms?

Tip: When working with exponential models, always ensure you understand the context and limitations, as real-world factors can affect the accuracy of the model.