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To solve this problem, let's start by understanding the differential equation and then use the given data to find the initial quantity of bacteria.

### Problem Summary
We are given:
1. The differential equation for bacterial growth:
   \[
   \frac{dx}{dt} = kx \quad \text{or} \quad \frac{dx}{x} = k \, dt
   \]
   where \( k \) is the proportionality factor.
2. Data points:
   - At \( t = 3 \) hours, \( x = 10^4 \).
   - At \( t = 5 \) hours, \( x = 4 \times 10^4 \).
3. We are asked to find the initial number of bacteria, \( x(0) \), at \( t = 0 \).

### Step 1: Solve the Differential Equation
Starting from the equation:
\[
\frac{dx}{x} = k \, dt
\]
Integrate both sides:
\[
\int \frac{dx}{x} = \int k \, dt
\]
This yields:
\[
\ln x = kt + C
\]
Exponentiating both sides, we get:
\[
x = e^{kt + C} = e^C \cdot e^{kt}
\]
Let \( e^C = x_0 \), where \( x_0 \) is the initial amount of bacteria. Thus:
\[
x = x_0 e^{kt}
\]

### Step 2: Use Given Data Points to Find \( k \) and \( x_0 \)
From the given information:
1. When \( t = 3 \), \( x = 10^4 \).
   \[
   10^4 = x_0 e^{3k}
   \]
2. When \( t = 5 \), \( x = 4 \times 10^4 \).
   \[
   4 \times 10^4 = x_0 e^{5k}
   \]

### Step 3: Formulate Equations to Solve for \( k \) and \( x_0 \)
Dividing the second equation by the first:
\[
\frac{4 \times 10^4}{10^4} = \frac{x_0 e^{5k}}{x_0 e^{3k}}
\]
Simplifying, we get:
\[
4 = e^{2k}
\]
Take the natural logarithm of both sides:
\[
\ln 4 = 2k
\]
\[
k = \frac{\ln 4}{2} = \ln 2
\]

### Step 4: Substitute \( k \) Back to Find \( x_0 \)
Now, use the first equation:
\[
10^4 = x_0 e^{3 \ln 2}
\]
Since \( e^{3 \ln 2} = 2^3 = 8 \), we have:
\[
10^4 = x_0 \cdot 8
\]
\[
x_0 = \frac{10^4}{8} = 1250
\]

### Conclusion
The initial number of bacteria was:
\[
x(0) = 1250
\]

### Answer
The correct choice is:
**B. \( \frac{10^4}{8} \) bacteria**

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Would you like a breakdown of any particular step or have further questions?

Here are some related questions:
1. How can we derive a general solution for exponential growth models?
2. What would change in the solution if the growth rate was not constant?
3. Can we use this model to predict the number of bacteria after 10 hours?
4. How do we interpret the constant \( k \) in real-world applications?
5. How would the solution differ if bacteria population followed a logistic growth model?

**Tip:** Exponential growth models are powerful in predicting population sizes when the growth rate remains constant over time.

Let x be the number of bacteria at a time t hours and x(0) be the number of bacteria at time t = 0. Given the differential equation dx/dt = kx or dx/x = kdt, where k is the proportionality factor. If there are approximately 10^4 bacteria at the end of 3 hours and 4 * 10^4 at the end of 5 hours, how many were there at the beginning?

This problem involves solving an exponential growth model for bacteria using a differential equation. Given specific values at different time points, determine the initial number of bacteria. Learn how to apply growth models and calculate the proportionality factor for real-world applications.

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