Exponential Continuous Growth
The exponential is commonly used to model natural processes of growth.
The formula of the exponential continuous growth is:
![P=P_o\cdot e^(kt)](https://img.qammunity.org/2023/formulas/mathematics/college/a3eu6lxma369b226ebq2ykdutpq8ahzaop.png)
Where:
Po = the initial population of the bacteria culture
k = a fixed constant
t = time
P = the population of the bacteria culture at any time t
The initial population of bacteria is given as Po=10
We are given the growth rate at 1.5% per minute.
The process starts at 7:00 am. At 7:01 am there will be 1.5% more bacteria than the previous minute, that is P=1.015*10=10.15 for t=1, thus:
![10.15=10\cdot e^(k(1))=10\cdot e^k](https://img.qammunity.org/2023/formulas/mathematics/college/8o613v4y4wa95gdxqxu8yr71qouf5ssyls.png)
Solving for k:
![\begin{gathered} 10\cdot e^k=10.15 \\ e^k=1.015 \\ k=\ln \text{ 1.015} \\ k=0.01489 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/sqtoklid9cnp9s9gfbn1cv1h88y1x6jdna.png)
Now we have the value of k, the function is:
![P(t)=10\cdot e^(0.01489t)](https://img.qammunity.org/2023/formulas/mathematics/college/6qyz1i3s52130kvk9f6uvzm6kxfojmz20k.png)
We are required to find the bacteria count after 12 hours. Since the time must be expressed in minutes, t=12 hours = 12*60 = 720 minutes
![P=10\cdot e^(0.01489\cdot720)=10\cdot45,242.9=452,429](https://img.qammunity.org/2023/formulas/mathematics/college/yr4qzalstw865hnmr6k2nvmsk3fou5bcx6.png)
The bacteria count will be 452,429 after 12 hours