Let's solve this step-by-step:

Part (a): Find the initial population

The exponential growth model can be written as: $P(t) = P_0 e^{kt}$ Where:

• $P(t)$: Population at time $t$
• $P_0$: Initial population
• $k$: Growth rate constant
• $t$: Time (in hours)

We are given:

• $P(2) = 125$
• $P(4) = 350$

Substituting these values into the formula:

1. For $t = 2$: 125 = P_0 e^{2k} \tag{1}

2. For $t = 4$: 350 = P_0 e^{4k} \tag{2}

Dividing (2) by (1) to eliminate $P_0$: $\frac{350}{125} = \frac{P_0 e^{4k}}{P_0 e^{2k}}$ $2.8 = e^{2k}$

Taking the natural logarithm ($\ln$): $\ln(2.8) = 2k$ $k = \frac{\ln(2.8)}{2} \approx 0.514$

Substitute $k$ into (1) to find $P_0$: $125 = P_0 e^{2(0.514)}$ $125 = P_0 e^{1.028}$ $P_0 = \frac{125}{e^{1.028}} \approx 45.01$

Initial population $P_0 \approx 45$.

Part (b): Write the exponential growth model

The exponential growth model is: $P(t) = 45 e^{0.514t}$

Part (c): Number of bacteria after 8 hours

Substitute $t = 8$ into the model: $P(8) = 45 e^{0.514 \times 8}$ $P(8) = 45 e^{4.112}$ $P(8) \approx 45 \times 61.11 = 2750$

The number of bacteria after 8 hours is approximately 2750.

Part (d): Time for bacteria to reach 25,000

We solve for $t$ when $P(t) = 25,000$: $25,000 = 45 e^{0.514t}$ $\frac{25,000}{45} = e^{0.514t}$ $555.56 = e^{0.514t}$

Taking the natural logarithm: $\ln(555.56) = 0.514t$ $t = \frac{\ln(555.56)}{0.514} \approx 13.7$

It will take approximately 13.7 hours for the bacteria count to reach 25,000.

If you need clarifications or further details, let me know!

Related Questions

1. How would the growth rate change if the population doubled in less time?
2. What would the initial population be if the growth rate were higher?
3. Can you derive the doubling time for this model?
4. How does this exponential growth compare to linear growth?
5. What happens to the population after 24 hours?

Tip

When dealing with exponential growth problems, always confirm if the given growth constant $k$ matches the units of time $t$.