Answer:
![P(t) = 100e^{0.0251t}](https://tex.z-dn.net/?f=P%28t%29%20%3D%20100e%5E%7B0.0251t%7D)
The doubling time is of 27.65 minutes.
Step-by-step explanation:
Exponential equation of growth:
The exponential equation for population growth is given by:
![P(t) = P(0)e^{kt}](https://tex.z-dn.net/?f=P%28t%29%20%3D%20P%280%29e%5E%7Bkt%7D)
In which P(0) is the initial value and k is the growth rate.
A freshly inoculated bacterial culture of Streptococcus contains 100 cells.
This means that
. So
![P(t) = 100e^{kt}](https://tex.z-dn.net/?f=P%28t%29%20%3D%20100e%5E%7Bkt%7D)
When the culture is checked 60 minutes later, it is determined that there are 450 cells present.
This means that
, and we use this to find k. So
![450 = 100e^{60k}](https://tex.z-dn.net/?f=450%20%3D%20100e%5E%7B60k%7D)
![e^{60k} = 4.5](https://tex.z-dn.net/?f=e%5E%7B60k%7D%20%3D%204.5)
![\ln{e^{60k}} = \ln{4.5}](https://tex.z-dn.net/?f=%5Cln%7Be%5E%7B60k%7D%7D%20%3D%20%5Cln%7B4.5%7D)
![60k = \ln{4.5}](https://tex.z-dn.net/?f=60k%20%3D%20%5Cln%7B4.5%7D)
![k = \frac{\ln{4.5}}{60}](https://tex.z-dn.net/?f=k%20%3D%20%5Cfrac%7B%5Cln%7B4.5%7D%7D%7B60%7D)
![k = 0.0251](https://tex.z-dn.net/?f=k%20%3D%200.0251)
So
![P(t) = 100e^{0.0251t}](https://tex.z-dn.net/?f=P%28t%29%20%3D%20100e%5E%7B0.0251t%7D)
Doubling time:
This is t for which P(t) = 2P(0) = 200. So
![200 = 100e^{0.0251t}](https://tex.z-dn.net/?f=200%20%3D%20100e%5E%7B0.0251t%7D)
![e^{0.0251t} = 2](https://tex.z-dn.net/?f=e%5E%7B0.0251t%7D%20%3D%202)
![\ln{e^{0.0251t}} = \ln{2}](https://tex.z-dn.net/?f=%5Cln%7Be%5E%7B0.0251t%7D%7D%20%3D%20%5Cln%7B2%7D)
![0.0251t = \ln{2}](https://tex.z-dn.net/?f=0.0251t%20%3D%20%5Cln%7B2%7D)
![t = \frac{\ln{2}}{0.0251}](https://tex.z-dn.net/?f=t%20%3D%20%5Cfrac%7B%5Cln%7B2%7D%7D%7B0.0251%7D)
![t = 27.65](https://tex.z-dn.net/?f=t%20%3D%2027.65)
The doubling time is of 27.65 minutes.
Here I'll Simplify it to show the work.
You need PEMDAS
9(2*2+4÷2)
9(4+4÷2)
9(8÷2)
9(4)
9(4) = 9 × 4 = 36
The Answer is 36.
Second option, just do Pemdas backwards.