Answer :
Given the problem, let's break it down step by step:
### Part (a)
We are given that the number of bacteria in a culture decreases according to a continuous exponential decay model. The initial population is 320 bacteria, and there are 240 bacteria left after 17 minutes.
1. Exponential Decay Model: The general model for exponential decay can be written as:
[tex]\[ y = y_0 e^{kt} \][/tex]
where [tex]\( y_0 \)[/tex] is the initial population, [tex]\( k \)[/tex] is the decay constant, and [tex]\( t \)[/tex] is time.
2. Initial Population: Here, [tex]\( y_0 = 320 \)[/tex].
3. Given Data Point:
After 17 minutes, the population is 240:
[tex]\[ 240 = 320 e^{k \cdot 17} \][/tex]
4. Solve for [tex]\( k \)[/tex]:
Rearrange the equation to solve for [tex]\( k \)[/tex]:
[tex]\[ \frac{240}{320} = e^{17k} \][/tex]
[tex]\[ 0.75 = e^{17k} \][/tex]
Taking the natural logarithm on both sides to solve for [tex]\( k \)[/tex]:
[tex]\[ \ln(0.75) = 17k \][/tex]
[tex]\[ k = \frac{\ln(0.75)}{17} \][/tex]
5. Plug [tex]\( k \)[/tex] back into the model:
Substituting [tex]\( k \)[/tex] into the exponential decay formula:
[tex]\[ y = 320 e^{\left( \frac{\ln(0.75)}{17} t \right)} \][/tex]
Simplifying further, using an exact representation of the expression:
[tex]\[ y = 320 \left( e^{\ln(0.75)} \right)^{\frac{t}{17}} \][/tex]
Since [tex]\( e^{\ln(a)} = a \)[/tex], we get:
[tex]\[ y = 320 \left(0.75^{\frac{t}{17}}\right) \][/tex]
Therefore, the formula relating [tex]\( y \)[/tex] to [tex]\( t \)[/tex] is:
[tex]\[ y = 320 \cdot 0.75^{\frac{t}{17}} \][/tex]
### Part (b)
To find the number of bacteria 21 minutes after the beginning of the study, substitute [tex]\( t = 21 \)[/tex] into the decay model.
1. Evaluation:
[tex]\[ y = 320 \cdot 0.75^{\frac{21}{17}} \][/tex]
Using the given expression:
[tex]\[ y = 320 \left( 0.75^{\frac{21}{17}} \right) \][/tex]
2. Compute [tex]\( y \)[/tex] at [tex]\( t = 21 \)[/tex]:
[tex]\[ y = 320 e^{\left( \frac{\ln(0.75)}{17} \cdot 21 \right)} \][/tex]
The rounded number of bacteria at [tex]\( t = 21 \)[/tex] minutes is:
[tex]\[ y \approx 224 \][/tex]
### Summary
- Formula relating [tex]\( y \)[/tex] to [tex]\( t \)[/tex]:
[tex]\[ y = 320 \cdot 0.75^{\frac{t}{17}} \][/tex]
- Number of bacteria at 21 minutes:
[tex]\[ 224 \text{ bacteria} \][/tex]
### Part (a)
We are given that the number of bacteria in a culture decreases according to a continuous exponential decay model. The initial population is 320 bacteria, and there are 240 bacteria left after 17 minutes.
1. Exponential Decay Model: The general model for exponential decay can be written as:
[tex]\[ y = y_0 e^{kt} \][/tex]
where [tex]\( y_0 \)[/tex] is the initial population, [tex]\( k \)[/tex] is the decay constant, and [tex]\( t \)[/tex] is time.
2. Initial Population: Here, [tex]\( y_0 = 320 \)[/tex].
3. Given Data Point:
After 17 minutes, the population is 240:
[tex]\[ 240 = 320 e^{k \cdot 17} \][/tex]
4. Solve for [tex]\( k \)[/tex]:
Rearrange the equation to solve for [tex]\( k \)[/tex]:
[tex]\[ \frac{240}{320} = e^{17k} \][/tex]
[tex]\[ 0.75 = e^{17k} \][/tex]
Taking the natural logarithm on both sides to solve for [tex]\( k \)[/tex]:
[tex]\[ \ln(0.75) = 17k \][/tex]
[tex]\[ k = \frac{\ln(0.75)}{17} \][/tex]
5. Plug [tex]\( k \)[/tex] back into the model:
Substituting [tex]\( k \)[/tex] into the exponential decay formula:
[tex]\[ y = 320 e^{\left( \frac{\ln(0.75)}{17} t \right)} \][/tex]
Simplifying further, using an exact representation of the expression:
[tex]\[ y = 320 \left( e^{\ln(0.75)} \right)^{\frac{t}{17}} \][/tex]
Since [tex]\( e^{\ln(a)} = a \)[/tex], we get:
[tex]\[ y = 320 \left(0.75^{\frac{t}{17}}\right) \][/tex]
Therefore, the formula relating [tex]\( y \)[/tex] to [tex]\( t \)[/tex] is:
[tex]\[ y = 320 \cdot 0.75^{\frac{t}{17}} \][/tex]
### Part (b)
To find the number of bacteria 21 minutes after the beginning of the study, substitute [tex]\( t = 21 \)[/tex] into the decay model.
1. Evaluation:
[tex]\[ y = 320 \cdot 0.75^{\frac{21}{17}} \][/tex]
Using the given expression:
[tex]\[ y = 320 \left( 0.75^{\frac{21}{17}} \right) \][/tex]
2. Compute [tex]\( y \)[/tex] at [tex]\( t = 21 \)[/tex]:
[tex]\[ y = 320 e^{\left( \frac{\ln(0.75)}{17} \cdot 21 \right)} \][/tex]
The rounded number of bacteria at [tex]\( t = 21 \)[/tex] minutes is:
[tex]\[ y \approx 224 \][/tex]
### Summary
- Formula relating [tex]\( y \)[/tex] to [tex]\( t \)[/tex]:
[tex]\[ y = 320 \cdot 0.75^{\frac{t}{17}} \][/tex]
- Number of bacteria at 21 minutes:
[tex]\[ 224 \text{ bacteria} \][/tex]
