Question

The count in a bacteria culture was 600 after 15 minutes and 1700 after 30 minutes.

A) What was the initial size of the culture?
B) Find the doubling period
C) Find the population after 70 minutes
D) When will the population reach 10000?

Any help would be appreciated, been at this for a while now.

Answer 1

r you sure you typed the problem correctly ?

Answer 2

100% sure

Answer 3

when do you need to submit this?

Answer 4

i got 2 hours of sleep time left so i will solve it after i get some sleep cant sit through the whole night, well almost did it already but still give me few hours and i will help u solve it

Answer 5

3 hours of sleep i mean

Answer 6

alright thanks, ill check up on it in the morning

Answer 7

hey i managed to get it after a while, no need to attempt the question. thanks anyways though

Answer 8

What did you get for your answers?

Answer 9

1,700 / 600 = 2.8333333333
So, every 15 minutes, the population increases by 2.8333333333

A) Therefore, the population at 0 minutes = 600 / 2.8333333333 = 212

B) The doubling period would be 'x' in the following equation:
2.83333333^x = 2
x * log(2.8333333) = log (2)
x = log(2) / log(2.8333333)
x = 0.3010299957 / 0.452297671
x = 0.6655572535 which needs to be multiplied by 15 minutes (the time interval in the problem)
0.6655572535 * 15 = 9.9833588022 minutes (the doubling period)

C) Population after 70 minutes
(70/15 = 4.6666666 time periods and the first population was measured after 15 minutes so we must calculate the time periods as 3.6666666)
The population after 70 minutes=
600 * 2.833333^3.6666666 = 27,326

D) Population reaches 10,000 when
600 * 2.833333^Y = 10,000
2.833333^Y = 10,000/600
2.833333^Y = 16.6666666667
Y*log(2.833333) = log(16.66666)
Y = log(16.666666) / log(2.8333333)
Y = 1.2218487496 / 0.452297671
Y = 2.7014261359
600 * 2.83333333^2.7014261359 does equal 10,000
The 2.7014261359 is the number of time periods so this has to be multiplied by 15
for the number of minutes
= 2.7014261359 * 15 =
40.52 minutes

Answer 10

I did it in a slightly more general fashion that doesn't require the population measurements to have been done at any particular interval (here they are conveniently evenly spaced).

\[P(t) = P_0 e^{-kt}\]where \(P_0\) is the initial population, and \(k\) is the exponential decay/growth constant.


We have \[P(15) = 600 = P_0 e^{-15k}\]and \[P(30) = 1700 = P_0e^{-30k}\]If we solve one equation for \(P_0\) we can substitute it into the other:
\[600 = P_0 e^{-15k}\]\[P_0 = 600 e^{+15k}\]\[1700 = 600e^{+15k}e^{-30k} = 600e^{-15k}\]Solving for \(k\),\[k =\frac{\ln(\frac{1700}{600})}{-15} \approx -0.06943\]

Now to find the population at \(P(0) = P_0\) we have \[ P_0 = 600 e^{(-0.06943)(15)}\approx 211.77\]from our substitution equation.

Doubling period can be gotten by solving \[2P_0 = P_0 e^{-kt_{2x}}\]\[2 = e^{-kt_{2x}}\]\[t_{2x} = \frac{ \ln(2)}{-k} \approx 9.983 \text{ minutes}\]

Population at 70 minutes is just a matter of evaluating\[P(70) = P_0e^{-k70} = (211.77)e^{-(-0.06943)*70} \approx 27326.7\]

And finally, finding the value of \(t\) where \(P(t) = 10000\):

\[10000 = P_0 e^{-kt_{10k}}\]\[\frac{\ln(\frac{10000}{P_0})}{-k} = t_{10k} \approx 55.521\text{ minutes}\]

@wolf1728 I didn't see where exactly your final calculation went astray, but here's a plot of \(P(t)\) with gridlines at \(t = 9.983, 55.521, 70\) showing the correctness of the figures above. Given that we agreed on the other answers, I think this plot is correct. Also a plot of the first 12 minutes more clearly showing the doubling time.

Answer 11

@wolf1728 Ah, I spotted the problem with your final answer. You started doubling from the population at t=15 (600), not t = 0 (211.77). Your answer is the amount of time elapsed after t=15 to get to population 10,000, and if we add your result to 15,
\[t_{10k} = 40.52+15 = 55.52\]in agreement with mine.

Answer 12

I figured that someone else besides you should figure out the answers just to have something as a comparison. Seems as if I made a mistake and good for you for finding it.