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The problem describes the exponential growth of bacteria, modeled by the differential equation:

\[
P'(t) = kP(t)
\]

This is a first-order linear differential equation where \( P(t) \) represents the number of bacteria at time \( t \) and \( k \) is the growth rate constant. 

### Step 1: Solve the differential equation

We recognize that this equation represents exponential growth, and its general solution is:

\[
P(t) = P_0 e^{kt}
\]

where \( P_0 \) is the initial number of bacteria (at \( t = 0 \)).

### Step 2: Use given data to find \( k \)

The problem provides two pieces of dataThe problem describes the exponential growth of bacteria, modeled by the differential equation:

\[
P'(t) = kP(t)
\]

This is a first-order linear differential equation where \( P(t) \) represents the number of bacteria at time \( t \) and \( k \) is the growth rate constant. 

### Step 1: Solve the differential equation

We recognize that this equation represents exponential growth, and its general solution is:

\[
P(t) = P_0 e^{kt}
\]

where \( P_0 \) is the initial number of bacteria (at \( t = 0 \)).

### Step 2: Use given data to find \( k \)

The problem provides two pieces of 
1. After 2 hours, \( P(2) = 200 \)
2. After 4 hours, \( P(4) = 800 \)

We will first use these two points to solve for the constant \( k \) and then find the initial value \( P_0 \).

At \( t = 2 \):
\[
P(2) = P_0 e^{2k} = 200
\]

At \( t = 4 \):
\[
P(4) = P_0 e^{4k} = 800
\]

### Step 3: Solve the system of equations

We now have the system:
\[
P_0 e^{2k} = 200
\]
\[
P_0 e^{4k} = 800
\]

Divide the second equation by the first to eliminate \( P_0 \):
\[
\frac{P_0 e^{4k}}{P_0 e^{2k}} = \frac{800}{200}
\]
\[
e^{2k} = 4
\]

Taking the natural logarithm of both sides:
\[
2k = \ln(4)
\]
\[
k = \frac{\ln(4)}{2} = \ln(2) \quad \text{(since \( \ln(4) = 2\ln(2) \))}.
\]

Thus, \( k = \ln(2) \).

### Step 4: Find \( P_0 \)

Substitute \( k = \ln(2) \) into the first equation:
\[
P_0 e^{2\ln(2)} = 200
\]
\[
P_0 \cdot 2^2 = 200
\]
\[
P_0 \cdot 4 = 200
\]
\[
P_0 = 50
\]

### Conclusion

The initial number of bacteria, \( P_0 \), is **50**.

---

Would you like further clarification or additional details on any part of this solution? Here are some related questions:

1. How do exponential growth models differ from linear growth models?
2. What are some real-world applications of differential equations in biology?
3. How would the solution change if there were a decay factor instead of growth?
4. What is the significance of the constant \( k \) in growth equations?
5. Can you derive a general formula for the time it takes for the population to double?

**Tip:** When solving exponential growth or decay problems, always look for easy ways to eliminate unknowns by dividing equations with similar terms.

The number P(t) of bacteria in a culture obeys the equation P'(t) = kP(t). After 2 hours, there were 200 bacteria present. After 4 hours, 800 bacteria were present. Determine the initial number of bacteria.

Learn how to determine the initial number of bacteria in a culture that grows exponentially over time. Given data at 2 hours and 4 hours, solve the differential equation P'(t) = kP(t) to find the initial population, suitable for Grades 10-12 and college students.

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