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The problem provides an exponential growth model \( P(t) = P_0 e^{kt} \), where:
- \( P(t) \) is the population of bacteria at time \( t \),
- \( P_0 \) is the initial population,
- \( k \) is the growth rate constant,
- \( t \) is time in minutes.

Given dataThe problem provides an exponential growth model \( P(t) = P_0 e^{kt} \), where:
- \( P(t) \) is the population of bacteria at time \( t \),
- \( P_0 \) is the initial population,
- \( k \) is the growth rate constant,
- \( t \) is time in minutes.

Given 
- \( P(20) = 700 \) (after 20 minutes, the population is 700),
- \( P(30) = 2000 \) (after 30 minutes, the population is 2000).

### Step 1: Set up the equations based on the given information
Using the exponential model:

1. For \( t = 20 \):
   \[
   700 = P_0 e^{20k}
   \]

2. For \( t = 30 \):
   \[
   2000 = P_0 e^{30k}
   \]

### Step 2: Solve for \( k \)
We can solve these two equations by dividing them:

\[
\frac{2000}{700} = \frac{P_0 e^{30k}}{P_0 e^{20k}}
\]

This simplifies to:

\[
\frac{2000}{700} = e^{10k}
\]

\[
\frac{2000}{700} = \frac{20}{7}
\]

So:

\[
\frac{20}{7} = e^{10k}
\]

Take the natural logarithm of both sides:

\[
\ln\left(\frac{20}{7}\right) = 10k
\]

Solving for \( k \):

\[
k = \frac{1}{10} \ln\left(\frac{20}{7}\right)
\]

### Step 3: Calculate \( k \)
Let me calculate this value.The growth rate constant \( k \approx 0.10498 \).

### Step 4: Solve for the initial population \( P_0 \)

We can substitute \( k \) into one of the original equations. Let's use \( P(20) = 700 \):

\[
700 = P_0 e^{20 \times 0.10498}
\]

Now, solve for \( P_0 \):

\[
P_0 = \frac{700}{e^{20 \times 0.10498}}
\]

Let me calculate \( P_0 \).The initial population \( P_0 \approx 85.75 \).

### Step 5: Answer for Part (b)

To find the population after 110 minutes, use the equation:

\[
P(110) = P_0 e^{110k}
\]

Let me calculate the population after 110 minutes.The population after 110 minutes is approximately 8,881,486.

### Step 6: Answer for Part (c)

To find the time when the population will reach 15,000, use the equation:

\[
15000 = P_0 e^{kt}
\]

Solving for \( t \):

\[
t = \frac{1}{k} \ln\left(\frac{15000}{P_0}\right)
\]

Let me calculate the time \( t \).It seems like I can’t do more advanced data analysis right now. Please try again later.

However, you can still calculate the time for the population to reach 15,000 using the following equation:

\[
t = \frac{1}{k} \ln\left(\frac{15000}{P_0}\right)
\]

Where:
- \( k \approx 0.10498 \),
- \( P_0 \approx 85.75 \).

If you'd like, I can walk you through the calculations step by step.

Would you like more details or further help? Here are 5 related questions:

1. How can you verify exponential growth models with different initial conditions?
2. How does changing the growth constant \( k \) affect the rate of growth?
3. What is the impact of rounding errors in exponential growth problems?
4. How can the exponential growth model be applied to other real-world scenarios?
5. How do you interpret the growth constant \( k \) in biological systems?

**Tip:** Always use as many decimal places as your system requires for precise results.

Assume that the number of bacteria follows an exponential growth model: P(t) = P_0e^{kt}. The count in the bacteria culture was 700 after 20 minutes and 2000 after 30 minutes. (a) What was the initial size of the culture? (b) Find the population after 110 minutes. (c) How many minutes after the start of the experiment will the population reach 15,000?

Learn how to solve an exponential growth problem involving a bacteria culture. Calculate the initial population, find the population after 110 minutes, and determine how many minutes it takes for the population to reach 15,000. This solution includes key algebraic and logarithmic formulas.

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