Question
The number of bacteria in a sample is increasing according to an exponential model. After two hours, the sample contained 800
bacteria. After six hours, the sample contained 3,200
bacteria. Write an exponential growth model for the number of bacteria in the sample after x
hours.
bacteria. After six hours, the sample contained 3,200
bacteria. Write an exponential growth model for the number of bacteria in the sample after x
hours.
Answers
Answered by
GPT-4o mini
To model the growth of the bacteria according to the exponential growth model, we can use the following general equation:
\[
N(t) = N_0 e^{kt}
\]
Where:
- \( N(t) \) is the number of bacteria at time \( t \),
- \( N_0 \) is the initial number of bacteria,
- \( k \) is the growth constant,
- \( t \) is the time in hours.
We are given two pieces of information:
1. After 2 hours, the sample contained 800 bacteria.
2. After 6 hours, the sample contained 3,200 bacteria.
Using these two data points, we can set up a system of equations.
### Step 1: Set up the equations
For \( t = 2 \):
\[
N(2) = N_0 e^{2k} = 800 \quad (1)
\]
For \( t = 6 \):
\[
N(6) = N_0 e^{6k} = 3200 \quad (2)
\]
### Step 2: Divide the equations to eliminate \( N_0 \)
Dividing equation (2) by equation (1) to eliminate \( N_0 \):
\[
\frac{N_0 e^{6k}}{N_0 e^{2k}} = \frac{3200}{800}
\]
This simplifies to:
\[
e^{4k} = 4
\]
### Step 3: Solve for \( k \)
Taking the natural logarithm of both sides:
\[
4k = \ln(4)
\]
Thus, we can find \( k \):
\[
k = \frac{\ln(4)}{4}
\]
### Step 4: Solve for \( N_0 \)
Now we can use \( k \) to find \( N_0 \). We can substitute \( k \) back into either equation (1) or (2). Let's use equation (1):
\[
N_0 e^{2k} = 800
\]
Substituting \( k \):
\[
N_0 e^{2(\frac{\ln(4)}{4})} = 800
\]
\[
N_0 e^{\frac{1}{2} \ln(4)} = 800
\]
\[
N_0 \cdot \sqrt{4} = 800
\]
\[
N_0 \cdot 2 = 800
\]
\[
N_0 = \frac{800}{2} = 400
\]
### Conclusion
Thus, we have \( N_0 = 400 \) and \( k = \frac{\ln(4)}{4} \).
### Final Exponential Growth Model
Substituting \( N_0 \) and \( k \) back into the general formula gives:
\[
N(t) = 400 e^{\frac{\ln(4)}{4}t}
\]
This can also be expressed more simply as:
\[
N(t) = 400 \cdot 4^{\frac{t}{4}}
\]
This is the exponential growth model for the number of bacteria in the sample after \( t \) hours.
\[
N(t) = N_0 e^{kt}
\]
Where:
- \( N(t) \) is the number of bacteria at time \( t \),
- \( N_0 \) is the initial number of bacteria,
- \( k \) is the growth constant,
- \( t \) is the time in hours.
We are given two pieces of information:
1. After 2 hours, the sample contained 800 bacteria.
2. After 6 hours, the sample contained 3,200 bacteria.
Using these two data points, we can set up a system of equations.
### Step 1: Set up the equations
For \( t = 2 \):
\[
N(2) = N_0 e^{2k} = 800 \quad (1)
\]
For \( t = 6 \):
\[
N(6) = N_0 e^{6k} = 3200 \quad (2)
\]
### Step 2: Divide the equations to eliminate \( N_0 \)
Dividing equation (2) by equation (1) to eliminate \( N_0 \):
\[
\frac{N_0 e^{6k}}{N_0 e^{2k}} = \frac{3200}{800}
\]
This simplifies to:
\[
e^{4k} = 4
\]
### Step 3: Solve for \( k \)
Taking the natural logarithm of both sides:
\[
4k = \ln(4)
\]
Thus, we can find \( k \):
\[
k = \frac{\ln(4)}{4}
\]
### Step 4: Solve for \( N_0 \)
Now we can use \( k \) to find \( N_0 \). We can substitute \( k \) back into either equation (1) or (2). Let's use equation (1):
\[
N_0 e^{2k} = 800
\]
Substituting \( k \):
\[
N_0 e^{2(\frac{\ln(4)}{4})} = 800
\]
\[
N_0 e^{\frac{1}{2} \ln(4)} = 800
\]
\[
N_0 \cdot \sqrt{4} = 800
\]
\[
N_0 \cdot 2 = 800
\]
\[
N_0 = \frac{800}{2} = 400
\]
### Conclusion
Thus, we have \( N_0 = 400 \) and \( k = \frac{\ln(4)}{4} \).
### Final Exponential Growth Model
Substituting \( N_0 \) and \( k \) back into the general formula gives:
\[
N(t) = 400 e^{\frac{\ln(4)}{4}t}
\]
This can also be expressed more simply as:
\[
N(t) = 400 \cdot 4^{\frac{t}{4}}
\]
This is the exponential growth model for the number of bacteria in the sample after \( t \) hours.
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