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:
[tex]N=N_0b^t[/tex]It is given that The count of bacteria was 100 after 15 minutes and 1300 after 30 minutes. So,
[tex]100=N_0b^{15}\text{ and 1300}=N_0b^{30}[/tex]Divide first equation by second to get:
[tex]\begin{gathered} \frac{100}{1300}=\frac{b^{15}}{b^{30}} \\ \frac{1}{13}=\frac{1}{b^{15}} \\ b^{15}=13 \\ b=(13)^{\frac{1}{15}} \end{gathered}[/tex]Put b = 1.186 in the first equation and solve for N_0:
[tex]\begin{gathered} 100=N_0(13^{\frac{1}{15}})^{15} \\ N_0=\frac{100}{13} \\ N_0=7.692 \end{gathered}[/tex]Thus, the initial size of the culture is 7.692.
To find doubling time, find the time taken to reach 200 count.
[tex]\begin{gathered} 200=7.962(13^{\frac{1}{15}})^t \\ \frac{200}{7.692}=(13)^{\frac{t}{15}} \\ 26=(13)^{\frac{t}{15}} \\ \ln26=\frac{t}{15}\ln(13) \\ \frac{t}{15}=\frac{\ln26}{\ln13} \\ t=19.053\text{ minutes} \end{gathered}[/tex]So, the time taken to double the count is 19.053 - 15 = 4.053 minutes.
The population after 115 minutes is given by:
[tex]\begin{gathered} N=7.692(13)^{\frac{115}{15}} \\ =2668527257.6037 \end{gathered}[/tex]Thus, the population after 115 minutes is 2668527257.6037.
The time taken to reach the population 15000 can be obtained as follows:
[tex]\begin{gathered} 15000=7.692(13)^{\frac{t}{15}} \\ \frac{15000}{7.692}=13^{\frac{t}{15}} \\ \ln(\frac{15000}{7.692})=\frac{t}{15}(\ln13) \\ t=44.302\text{ minutes} \end{gathered}[/tex]Thus, the population reaches 15000 after 44.302 minutes.
