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We are given a logistic growth model for yeast growth, with the following parameters:

- **Limiting population (carrying capacity)** \( m = 3.7 \times 10^8 \),
- **Growth rate constant** \( k = 0.55 \),
- **Two population values**: 
  - \( P_1 = 2.1 \times 10^8 \) (initial population at a given time),
  - \( P_2 = 2.6 \times 10^8 \) (larger population at the same time).

The problem asks two things:
1. How much time was required for the first population to reach the size of the second?
2. What was the growth rate at that time?

### 1. Logistic Growth Model

The logistic growth equation is given by:

\[
P(t) = \frac{m}{1 + \left(\frac{m - P_0}{P_0}\right) e^{-kt}},
\]
where:
- \( P(t) \) is the population at time \( t \),
- \( P_0 \) is the initial population,
- \( m \) is the limiting population (carrying capacity),
- \( k \) is the growth rate constant,
- \( t \) is time.

We are tasked to find the time required for \( P_1 = 2.1 \times 10^8 \) to grow to \( P_2 = 2.6 \times 10^8 \).

### Step 1: Calculate Time to Reach \( P_2 \)

We know:
\[
P_1 = 2.1 \times 10^8 \quad \text{and} \quad P_2 = 2.6 \times 10^8.
\]

We can rearrange the logistic growth equation to solve for time \( t \):

\[
P(t) = \frac{m}{1 + \left(\frac{m - P_1}{P_1}\right) e^{-kt}}.
\]

At \( t = t_2 \), we have:

\[
P_2 = \frac{m}{1 + \left(\frac{m - P_1}{P_1}\right) e^{-kt}}.
\]

Substituting \( P_2 = 2.6 \times 10^8 \), \( P_1 = 2.1 \times 10^8 \), \( m = 3.7 \times 10^8 \), and \( k = 0.55 \), we solve for \( t \).

### Step 2: Growth Rate at Time \( t \)

The rate of change of the population is given by:

\[
\frac{dP}{dt} = k P \left( 1 - \frac{P}{m} \right).
\]

We will calculate this at \( P = 2.1 \times 10^8 \).

Let me calculate these steps.1. **Time required to grow from \( P_1 = 2.1 \times 10^8 \) to \( P_2 = 2.6 \times 10^8 \)**: It takes approximately **1.07 hours**.

2. **Growth rate at the time the population was \( P_1 = 2.1 \times 10^8 \)**: The growth rate at that moment was approximately **\( 4.99 \times 10^7 \) cells per hour**.

Would you like further details or clarifications?

Here are 5 related questions for deeper understanding:
1. How would the time change if the growth rate \( k \) were lower?
2. How is the growth rate influenced by the current population size in logistic growth?
3. What happens to the growth rate as the population approaches the carrying capacity \( m \)?
4. Could you derive the general time formula for any two populations in logistic growth?
5. How does logistic growth differ from exponential growth in terms of population dynamics?

**Tip**: In logistic growth, the population grows fastest when it's halfway to the carrying capacity (i.e., \( P = m/2 \)).

A population of yeast Saccharomyces cere-vislae was put in a flask with nutrients, found to grow logistically with a limiting population of 3.7 × 10^8 and a growth rate of 0.55. One population was found to be 2.1 × 10^8, and another at 2.6 × 10^8. How much time was required for the first population to reach the second size? How fast was it growing at that time?

Explore a logistic growth problem involving yeast population Saccharomyces cere-vislae, with a limiting population of 3.7 × 10^8 and a growth rate of 0.55. Find the time for population growth from 2.1 × 10^8 to 2.6 × 10^8 and the rate of growth at that time.

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