Given:
The population of the bacteria at the beginning = 80
the population grows according to a continuous exponential growth model.
after 14 days, there are 216 bacteria.
y = the number of bacteria after time t
So, the general relation between y and t will be:
![y=a\cdot e^(bt)](https://img.qammunity.org/2023/formulas/mathematics/high-school/fb9hkfwct81z3e0xcndo8bk04lhmx2pjmi.png)
We need to find the values of a and b
At t = 0 y = 80
So,
![\begin{gathered} 80=a\cdot e^0 \\ a=80 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/high-school/aamv0q165dvqhinaohu3ovmo4zg2n2bqkr.png)
When t = 14 , y = 216
So,
![\begin{gathered} 216=80e^(14b) \\ (216)/(80)=e^(14b) \\ \text{2}.7=e^(14b) \\ \ln 2.7=14b \\ \text{0}.99325=14b \\ b=(0.99325)/(14)=0.071 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/high-school/dqtbgq693cc46pxa9jooskdcetjpi0be7v.png)
so, the function will be:
![y=80\cdot e^(0.071t)](https://img.qammunity.org/2023/formulas/mathematics/high-school/wwuzscd0qmrgnzwsjswa4mh9okedt29yne.png)
Part b: we need to find the number of bacteria after 23 days
So, substitute with t = 23
so,
![y=80\cdot e^(0.071\cdot23)=409](https://img.qammunity.org/2023/formulas/mathematics/high-school/lbbbyrxsg2gybl1gzaipugqcdch1rexuur.png)
so, after 23 days the number of bacteria = 409