Answer:
The function that can be used to describe the number (n) of bacteria after 2 minutes is;
![P = 4 \cdot e^{\left(\dfrac{ln(32)}{5} \times 2\right)} \approx 4 \cdot e^{\left(0.693\times 2\right)}](https://tex.z-dn.net/?f=P%20%3D%204%20%5Ccdot%20e%5E%7B%5Cleft%28%5Cdfrac%7Bln%2832%29%7D%7B5%7D%20%5Ctimes%202%5Cright%29%7D%20%5Capprox%204%20%5Ccdot%20e%5E%7B%5Cleft%280.693%5Ctimes%202%5Cright%29%7D)
Step-by-step explanation:
The data in the table are presented as follows;
Number of bacteria; 4, 128, 4,096, 131,072
Number of minutes from initial state; 0, 5, 10, 15
The general equation for population growth is presented as follows;
![P = P_0 \cdot e^{r\cdot t}](https://tex.z-dn.net/?f=P%20%3D%20P_0%20%5Ccdot%20e%5E%7Br%5Ccdot%20t%7D)
Where;
P = The population after 't' minutes
P₀ = The initial population
r = The population growth rate
t = The time taken for the growth in population numbers
At t = minutes. we have;
![4 = P_0 \cdot e^{r\times0} = P_0](https://tex.z-dn.net/?f=4%20%3D%20P_0%20%5Ccdot%20e%5E%7Br%5Ctimes0%7D%20%3D%20P_0)
∴ P₀ = 4
At t = 5, we have;
![128 = 4 \cdot e^{r\times 5}](https://tex.z-dn.net/?f=128%20%3D%204%20%5Ccdot%20e%5E%7Br%5Ctimes%205%7D)
![\therefore e^{r\times 5} = \dfrac{128}{4} = 32](https://tex.z-dn.net/?f=%5Ctherefore%20%20e%5E%7Br%5Ctimes%205%7D%20%3D%20%5Cdfrac%7B128%7D%7B4%7D%20%3D%2032)
![ln\left(e^{r\times 5}\right) = ln(32)](https://tex.z-dn.net/?f=ln%5Cleft%28e%5E%7Br%5Ctimes%205%7D%5Cright%29%20%3D%20ln%2832%29)
∴ r × 5 = ㏑(32)
r = ln(32)/5 ≈ 0.693
The number (n) of bacteria after 2 minutes is therefore;
![P = 4 \cdot e^{\left(\dfrac{ln(32)}{5} \times 2\right)} \approx 4 \cdot e^{\left(0.693\times 2\right)}](https://tex.z-dn.net/?f=P%20%3D%204%20%5Ccdot%20e%5E%7B%5Cleft%28%5Cdfrac%7Bln%2832%29%7D%7B5%7D%20%5Ctimes%202%5Cright%29%7D%20%5Capprox%204%20%5Ccdot%20e%5E%7B%5Cleft%280.693%5Ctimes%202%5Cright%29%7D)