The problem describes a classic case of exponential growth, where the rate of population increase is proportional to the current population size. The model for exponential growth can be expressed as:

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

Where:

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

We are given two conditions:

1. After 3 hours, $P(3) = 500$,
2. After 10 hours, $P(10) = 3000$.

We need to find $P_0$, the initial population.

Step 1: Set up the equations

Using the general form of the equation $P(t) = P_0 e^{kt}$, plug in the given values to set up two equations:

For $t = 3$: $500 = P_0 e^{3k} \quad \text{(Equation 1)}$

For $t = 10$: $3000 = P_0 e^{10k} \quad \text{(Equation 2)}$

Step 2: Solve for $k$

Divide Equation 2 by Equation 1 to eliminate $P_0$: $\frac{3000}{500} = \frac{P_0 e^{10k}}{P_0 e^{3k}}$ This simplifies to: $6 = e^{7k}$ Take the natural logarithm (ln) of both sides: $\ln(6) = 7k$ $k = \frac{\ln(6)}{7}$

Step 3: Solve for $P_0$

Now that we have $k$, substitute it back into Equation 1 to solve for $P_0$: $500 = P_0 e^{3k}$ Substitute $k = \frac{\ln(6)}{7}$ into this equation: $500 = P_0 e^{3 \times \frac{\ln(6)}{7}}$ This simplifies to: $500 = P_0 e^{\frac{3}{7} \ln(6)}$ Use the property $e^{a \ln(b)} = b^a$: $500 = P_0 \times 6^{\frac{3}{7}}$ Now solve for $P_0$: $P_0 = \frac{500}{6^{\frac{3}{7}}}$

I'll calculate the final value of $P_0$.The initial number of bacteria, $P_0$, is approximately 232.

Would you like further clarification on any of the steps? Here are some related questions to deepen your understanding:

1. What does it mean for the growth rate to be proportional to the population?
2. How can we model exponential decay instead of growth?
3. Why do we use natural logarithms in solving exponential equations?
4. How would the model change if the population reached a carrying capacity?
5. What are other real-world situations that can be modeled using exponential growth?

Tip: Always double-check your work by substituting the calculated values back into the original equations to verify correctness.