Given two points for an exponential function,

Use the two points to find the growth rate, k. Write an exponential model for each point, then solve this system of two equations for k.
Use either point with the k you found to find the initial amount at time zero, Ao.
Doubling time is when the amount is 2*Ao.
Use the values of k and Ao to calculate the amount for a given time or to find the time to reach a specific amount in the future.
The count in a bacteria culture was 200 after 15 minutes and 1100 after 40 minutes. Assuming the count grows exponentially. You may enter the exact value or round to 2 decimal places.

What was the initial size of the culture?

Find the doubling period.

Find the population after 110 minutes.

When will the population reach 12000.

To solve this problem, we'll break it down step by step.

### 1. **Find the growth rate \( k \):**
Given two points \((t_1, A_1)\) and \((t_2, A_2)\) on the exponential curve, the exponential model can be written as:

\[
A(t) = A_0 \cdot e^{kt}
\]

For the given points:
- Point 1: \( (15, 200) \)
- Point 2: \( (40, 1100) \)

We can set up two equations:

\[
200 = A_0 \cdot e^{15k}
\]
\[
1100 = A_0 \cdot e^{40k}
\]

To eliminate \(A_0\), divide the second equation by the first:

\[
\frac{1100}{200} = \frac{A_0 \cdot e^{40k}}{A_0 \cdot e^{15k}}
\]

Simplifying:

\[
5.5 = e^{25k}
\]

Take the natural logarithm on both sides to solve for \( k \):

\[
\ln(5.5) = 25k
\]

\[
k = \frac{\ln(5.5)}{25}
\]

Let's calculate \( k \):

\[
k \approx \frac{1.7047}{25} \approx 0.06819
\]

### 2. **Find the initial amount \( A_0 \):**
Now, substitute \( k \) back into one of the original equations, say \( 200 = A_0 \cdot e^{15k} \):

\[
200 = A_0 \cdot e^{15 \times 0.06819}
\]

Calculate the exponential term:

\[
e^{15 \times 0.06819} \approx e^{1.02285} \approx 2.7812
\]

So,

\[
A_0 = \frac{200}{2.7812} \approx 71.91
\]

The initial size of the culture \( A_0 \) was approximately **71.91** bacteria.

### 3. **Find the doubling period:**
The doubling period \( T_d \) satisfies:

\[
2A_0 = A_0 \cdot e^{kT_d}
\]

\[
2 = e^{kT_d}
\]

Taking the natural logarithm:

\[
\ln(2) = kT_d
\]

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

Substitute the value of \( k \):

\[
T_d \approx \frac{0.6931}{0.06819} \approx 10.16 \text{ minutes}
\]

The doubling period is approximately **10.16 minutes**.

### 4. **Find the population after 110 minutes:**
Use the exponential model \( A(t) = A_0 \cdot e^{kt} \):

\[
A(110) = 71.91 \cdot e^{0.06819 \times 110}
\]

Calculate the exponent:

\[
e^{0.06819 \times 110} \approx e^{7.501} \approx 1809.14
\]

So,

\[
A(110) \approx 71.91 \times 1809.14 \approx 130021.42
\]

The population after 110 minutes is approximately **130,021** bacteria.

### 5. **When will the population reach 12,000?**
We need to solve for \( t \) in:

\[
12000 = 71.91 \cdot e^{0.06819t}
\]

Divide both sides by 71.91:

\[
\frac{12000}{71.91} = e^{0.06819t}
\]

\[
166.87 \approx e^{0.06819t}
\]

Take the natural logarithm on both sides:

\[
\ln(166.87) \approx 0.06819t
\]

\[
t \approx \frac{\ln(166.87)}{0.06819} \approx \frac{5.1185}{0.06819} \approx 75.07 \text{ minutes}
\]

The population will reach 12,000 bacteria in approximately **75.07 minutes**.

---

### Summary:
- Initial size of the culture: **71.91** bacteria
- Doubling period: **10.16** minutes
- Population after 110 minutes: **130,021** bacteria
- Time to reach 12,000 bacteria: **75.07** minutes

Do you want more details or have any questions?

### Relative Questions:
1. How do you derive the formula for the growth rate \( k \) from two points on an exponential curve?
2. What are the key differences between linear and exponential growth?
3. How does the doubling period change if the growth rate \( k \) increases?
4. What would be the impact on the initial size if the first point was taken at a different time?
5. How can we use logarithms to simplify exponential equations in different scenarios?

### Tip:
Remember, for exponential growth problems, identifying the growth rate \( k \) and the initial amount \( A_0 \) from given data points is essential for predicting future values or determining when a certain population size will be reached.

Learn how to solve problems involving exponential growth models with specific examples. Find the initial size of a population, determine the doubling period, and predict future population sizes based on given data points. This problem involves calculating growth rates, initial amounts, and exponential equations using practical examples. Suitable for advanced high school students and beyond.

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