Question

I got closed out of myLast session for “connection issues”. Thanks for the help in advance !

Answer 1

Given:

The count of bacteria was 100 after 15 minutes and 1300 after 30 minutes.

To find:

1. The initial size of the culture.

2. The doubling period.

3. Find the population after 115 minutes.

4. The time at which population 15000.

Solution:

The growth is exponential. Let the equation that shows the bacteria after t minutes is given by:

`N=N_0b^t`

It is given that The count of bacteria was 100 after 15 minutes and 1300 after 30 minutes. So,

`100=N_0b^(15)\text{ and 1300}=N_0b^(30)`

Divide first equation by second to get:

`\begin{gathered} (100)/(1300)=(b^(15))/(b^(30)) \\ (1)/(13)=(1)/(b^(15)) \\ b^(15)=13 \\ b=(13)^{(1)/(15)} \end{gathered}`

Put b = 1.186 in the first equation and solve for N_0:

`\begin{gathered} 100=N_0(13^{(1)/(15)})^(15) \\ N_0=(100)/(13) \\ N_0=7.692 \end{gathered}`

Thus, the initial size of the culture is 7.692.

To find doubling time, find the time taken to reach 200 count.

`\begin{gathered} 200=7.962(13^{(1)/(15)})^t \\ (200)/(7.692)=(13)^{(t)/(15)} \\ 26=(13)^{(t)/(15)} \\ \ln26=(t)/(15)\ln(13) \\ (t)/(15)=(\ln26)/(\ln13) \\ t=19.053\text{ minutes} \end{gathered}`

So, the time taken to double the count is 19.053 - 15 = 4.053 minutes.

The population after 115 minutes is given by:

`\begin{gathered} N=7.692(13)^{(115)/(15)} \\ =2668527257.6037 \end{gathered}`

Thus, the population after 115 minutes is 2668527257.6037.

The time taken to reach the population 15000 can be obtained as follows:

`\begin{gathered} 15000=7.692(13)^{(t)/(15)} \\ (15000)/(7.692)=13^{(t)/(15)} \\ \ln((15000)/(7.692))=(t)/(15)(\ln13) \\ t=44.302\text{ minutes} \end{gathered}`

Thus, the population reaches 15000 after 44.302 minutes.