The problem describes bacterial population growth following a continuous exponential growth model with a growth rate of 6% per hour.

The formula for continuous exponential growth is:

$P(t) = P_0 \cdot e^{rt}$

Where:

• $P(t)$ is the population at time $t$,
• $P_0$ is the initial population,
• $r$ is the growth rate (as a decimal),
• $t$ is the time in hours,
• $e$ is Euler's number (approximately 2.71828).

Given dataThe problem describes bacterial population growth following a continuous exponential growth model with a growth rate of 6% per hour.

The formula for continuous exponential growth is:

$P(t) = P_0 \cdot e^{rt}$

Where:

• $P(t)$ is the population at time $t$,
• $P_0$ is the initial population,
• $r$ is the growth rate (as a decimal),
• $t$ is the time in hours,
• $e$ is Euler's number (approximately 2.71828).

Given

• Initial population $P_0 = 265$ bacteria,
• Growth rate $r = 0.06$ (since 6% = 0.06),
• Time $t = 4$ hours.

Now, we will substitute these values into the formula to find the population after 4 hours:

$P(4) = 265 \cdot e^{0.06 \cdot 4}$

First, calculate the exponent: $0.06 \cdot 4 = 0.24$

Next, calculate $e^{0.24}$: $e^{0.24} \approx 1.27125$

Finally, calculate the population: $P(4) = 265 \cdot 1.27125 \approx 336.9$

So, the population after 4 hours is approximately 336.9 bacteria.

Would you like further details or explanations on any part of this calculation?

Here are 5 related questions for deeper understanding:

1. How does the population change if the growth rate increases to 10%?
2. What happens to the population after 6 hours under the same growth model?
3. Can the exponential model be used for other biological growth processes, and if so, how?
4. What is the significance of Euler's number $e$ in continuous growth models?
5. How would the solution change if the population were modeled using a logistic growth curve?

Tip: In continuous exponential growth, the population grows at a rate proportional to its current size, which means growth accelerates over time.