If the number of bacteria in a colony doubles every 236 minutes and there is currently a population of 9,085 bacteria, what will the population be 708 minutes from now?

Question

asked Jul 6, 2023 188k views

1 Answer

Visibleman · Answer 1 · 2023-07-13T02:10:51+0000

We could state an exponential model.

Number of bateria will be given by:

where k=growth constant and t=time in minutes.

We're given that N0 is the initial population, which is:

We also know that, when t=236 min, N=2(9,085) = 18,170:

$18,170=9,085e^(k(236))^{}$

We're going to solve this equation to find the value of k:

$\begin{gathered} (18,170)/(9,085)=e^(236k) \\ 2=e^(236k) \\ \ln 2=\ln e^(236k) \\ \ln 2=236k \\ k=(\ln 2)/(236) \end{gathered}$

Then, our expression is:

$N=9,085e^{(\ln 2)/(236)t}$

To find the population of bacteria after 708 minutes, we replace t by 708:

$N=9085e^{(\ln2)/(236)(708)}=72680$

Therefore, the population of bacteria 708 minutes from now, is 72680.