When initially observed, a container of bacteria under investigation has a total of 123 cells. For five days the growth of the cells has been continuous and has reached a count of 440. Assume the growth rate is constant and continuous.

a) What is the growth rate K of the cells?

b) How many cells would have been present two days before the initial observation?

c) How many days from the initial observation will it take for the cells to reach 10,000?

Answers:
a) K= .255
b) y= 73.86 cells
c) t= 17.25 days

How do i derrive at these answers. I have tried my best to find a connection between the problem and the answers but I have not yet been able to find one. Any help would be of great appreciation.

Sincerely,
Hawk 789

A = Pe^kt

...where "A" is the ending amount, "P" is the beginning amount, "k" is the growth (or decay) constant, and "t" is the time.

In your case, you have P = 123, and A = 440 at t = 5. Using this, you can solve for the constant k:

440 = 123e^5k

440/123 = e^5k

ln(440/123) = 5k

ln(440/123)/5 = k

(Growth/decay constants are often messy like this. Do NOT simplify or approximate with a truncated decimal! Round-off error can quickly become huge with logs and exponentials, so stay "exact" for as long as you can, trying only to round at the very end!)

To figure the number of cells at t = -2, plug in -2 for t and ln(440/123)/5 for k, along with 123 for P, and simplify. I get approximately 74. (Remember that you can't have fractions of cells, so you'll have to round to a whole number to get a REASONABLE answer; their answer is not physically reasonable.)

For the last part, plug in ln(440/123)/5 for k, 123 for P, and 10000 for A. Solve for the time t. (Decimals/fractions are okay here.)

Hope that helps!