The count in a bacteria culture was 200 after 20 minutes and 1900 after 35 minutes. Assuming the count grows exponentially, What was the initial size of the culture? Find the doubling period. Find the population after 120 minutes. When will the population reach 11000. You may enter the exact value or round to 2 decimal places.


We are given that the growth of the bacteria culture is exponential. The general form for exponential growth is:

\[
P(t) = P_0 e^{kt}
\]
where:
- \( P(t) \) is the population at time \( t \),
- \( P_0 \) is the initial population (what we're solving for),
- \( k \) is the growth rate constant,
- \( t \) is the time in minutes,
- \( e \) is Euler's number (approximately 2.718).

### Step 1: Set up the equations from the given information
We know the following:
- At \( t = 20 \) minutes, the population is 200: 
  \[
  P(20) = 200 = P_0 e^{20k}
  \]
- At \( t = 35 \) minutes, the population is 1900:
  \[
  P(35) = 1900 = P_0 e^{35k}
  \]

### Step 2: Solve for \( k \)
To find \( k \), divide the second equation by the first to eliminate \( P_0 \):

\[
\frac{1900}{200} = \frac{P_0 e^{35k}}{P_0 e^{20k}}
\]

\[
9.5 = e^{15k}
\]

Now take the natural logarithm of both sides:

\[
\ln(9.5) = 15k
\]

\[
k = \frac{\ln(9.5)}{15} \approx \frac{2.2513}{15} \approx 0.15009
\]

### Step 3: Solve for \( P_0 \)
Substitute \( k \approx 0.15009 \) into one of the original equations to find \( P_0 \). Let's use the equation for \( t = 20 \):

\[
200 = P_0 e^{20(0.15009)}
\]

\[
200 = P_0 e^{3.0018}
\]

\[
P_0 = \frac{200}{e^{3.0018}} \approx \frac{200}{20.1} \approx 9.95
\]

So, the initial population \( P_0 \) is approximately **9.95**.

### Step 4: Find the doubling period
The doubling period \( T_d \) is the time it takes for the population to double. It can be calculated using:

\[
T_d = \frac{\ln(2)}{k}
\]

Substitute \( k \approx 0.15009 \):

\[
T_d = \frac{\ln(2)}{0.15009} \approx \frac{0.6931}{0.15009} \approx 4.62 \text{ minutes}
\]

### Step 5: Find the population after 120 minutes
Use the formula \( P(t) = P_0 e^{kt} \) to find the population at \( t = 120 \):

\[
P(120) = 9.95 e^{0.15009 \times 120}
\]

\[
P(120) = 9.95 e^{18.0108} \approx 9.95 \times 660807.3 \approx 6570954.67
\]

So, the population after 120 minutes is approximately **6,570,954.67**.

### Step 6: Find when the population will reach 11,000
To find when the population reaches 11,000, we solve:

\[
11000 = 9.95 e^{0.15009t}
\]

Divide both sides by 9.95:

\[
1105.53 = e^{0.15009t}
\]

Take the natural logarithm of both sides:

\[
\ln(1105.53) = 0.15009t
\]

\[
t = \frac{\ln(1105.53)}{0.15009} \approx \frac{7.008}{0.15009} \approx 46.68 \text{ minutes}
\]

So, the population will reach 11,000 after approximately **46.68 minutes**.

---

### Summary of Results:
1. **Initial population \( P_0 \)**: 9.95
2. **Doubling period \( T_d \)**: 4.62 minutes
3. **Population after 120 minutes**: 6,570,954.67
4. **Time to reach a population of 11,000**: 46.68 minutes

---

Would you like to explore any of these calculations further? Here are 5 related questions:

1. How does the growth constant \( k \) affect the doubling period?
2. Can we use this method for other exponential growth scenarios, like radioactive decay?
3. What would the population be after 240 minutes?
4. What if the bacteria grew at a slower rate, say half the current growth constant?
5. How would a change in the initial population affect the results?

**Tip**: When solving exponential growth problems, always check if you can simplify by taking logarithms at the right step!

The count in a bacteria culture was 200 after 20 minutes and 1900 after 35 minutes. Assuming the count grows exponentially, What was the initial size of the culture? Find the doubling period. Find the population after 120 minutes. When will the population reach 11000. You may enter the exact value or round to 2 decimal places.

Explore how to calculate the initial size of a bacteria culture, find the doubling period, and predict the population after 120 minutes using exponential growth formulas. Detailed steps included for solving exponential growth problems, suitable for Grades 10-12 students.

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