The number of bacteria in a dish has tripled every month. If there were 2500 bacteria at first, how many bacte

Question

Please help :(

I don't know how to set up an equation for this. I'm stumped. Please help

Answer 1

Bacteria growth rate = k
Bacteria = b
Time = t

Rate of growth of bacteria is proportional to existing number of bacteria
Mathematically,
db/dt = k times b = kb

Integrating both sides
b = Ae^(kt) + B (A and B are constants of integration) .... (1)
At t = 0, b = 2500
Plugging >> 2500 = A + B
At t = 30 days = 30 x 24 x 60 x 60 seconds, b = 3 x 2500
7500=Ae^(k times 720 x 3600) + B

For simplification, you may assume B = 0 and solve for A and k
A = 2500

7500 = 2500 e^(k x 720 x 3600)
k = ln(3)/(720 x 3600)

Plug values of A, B, k into equation (1), and sail

Answer 2

Sometimes people read too much into a problem, causing them much unnecessary trouble. It seems that this may be the case here for some of us.

This is merely a straightforward case of geometric growth. Nowhere is it stated that the rate of growth is proportional to the number of bacteria in the dish, only that it triples from month to month. So all the application of calculus techniques is entirely unneeded.

The problem states that the bacteria population triples from one month to the next over a period of three months. If we let p(0) be the original population, then the population after one month, p(1), is 3p(0). The population after two months, p(2), is 3p(1), and at the end of three months, p(3), is 3p(2).

So the equation for the final population, p(3), is

p(3) = 3(3(3p(0))) = 3^3p(0) = 27p(0).

So p(3) is 27 x 2500 = 67,500.

Answer 3

N=2500*3^m

where N=number of bacteria
m=number of months