Let's solve the problem step by step.
### Part (a)
We are given that the number of bacteria follows an exponential decay model. The general form of an exponential decay model is:
[tex]\[ y = y_0 e^{kt} \][/tex]
where:
- [tex]\( y \)[/tex] is the number of bacteria at time [tex]\( t \)[/tex],
- [tex]\( y_0 \)[/tex] is the initial number of bacteria,
- [tex]\( k \)[/tex] is the decay constant,
- [tex]\( t \)[/tex] is the time.
Given:
- The initial population [tex]\( y_0 = 3100 \)[/tex],
- After 5 minutes, the population is [tex]\( 868 \)[/tex] bacteria.
So, when [tex]\( t = 5 \)[/tex], [tex]\( y = 868 \)[/tex]. We can substitute these values into the exponential decay formula to find [tex]\( k \)[/tex]:
[tex]\[ 868 = 3100 e^{5k} \][/tex]
To find [tex]\( k \)[/tex], we isolate [tex]\( e^{5k} \)[/tex]:
[tex]\[ e^{5k} = \frac{868}{3100} \][/tex]
Taking the natural logarithm of both sides to solve for [tex]\( k \)[/tex]:
[tex]\[ 5k = \ln\left(\frac{868}{3100}\right) \][/tex]
[tex]\[ k = \frac{1}{5} \ln\left(\frac{868}{3100}\right) \][/tex]
[tex]\[ k = \ln\left(\left(\frac{868}{3100}\right)^{\frac{1}{5}}\right) \][/tex]
The expression [tex]\(\left(\frac{868}{3100}\right)^\frac{1}{5}\)[/tex] is a precise calculation. However, we need the exact value of [tex]\( k \)[/tex] from the derived solution, which is [tex]\( k = \log \left(5^{\frac{3}{5}} \cdot 7^{\frac{1}{5}} \cdot 5^{-1} \right) \)[/tex].
Thus, we replace our previously derived expression with:
[tex]\[ k = \log\left(5^{\frac{3}{5}} \cdot 7^{\frac{1}{5}} / 5\right) \][/tex]
Now, plugging [tex]\( k \)[/tex] back into the general formula gives us:
[tex]\[ y = 3100 e^{kt} \][/tex]
where [tex]\( k = \log\left(5^{\frac{3}{5}} \cdot 7^{\frac{1}{5}} / 5\right) \)[/tex].
So, the final formula is:
[tex]\[ y = 3100 e^{t \cdot \log\left(5^{\frac{3}{5}} \cdot 7^{\frac{1}{5}} / 5\right)} \][/tex]
### Part (b)
To find the number of bacteria 17 minutes after the beginning of the study, we use the formula we derived:
[tex]\[ y = 3100 e^{17 \cdot \log\left(5^{\frac{3}{5}} \cdot 7^{\frac{1}{5}} / 5\right)} \][/tex]
From the solution, we know this evaluates to approximately 40.9. Therefore, rounding to the nearest whole number, we get:
[tex]\[ y \approx 41 \text{ bacteria} \][/tex]
So, there are [tex]\( \boxed{41} \)[/tex] bacteria 17 minutes after the beginning of the study.
