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A certain strain of bacteria that is growing on your kitchen counter doubles every 5 minutes. Assuming you start with only one bacterium, how many bacteria could be present at the end of 96 minutes?

I've tried $\dfrac{96}{5}$ because "every 5 minutes", but I believe we are supposed to use Euler's constant, $e$, along with a certain formula to measure exponential growth. I just can't think of the formula to use.

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After 96 minutes, the bacteria have doubled $\lfloor 96/5\rfloor=19$ times and are still working on the 20th generation. What do you get if you start with 1 and double it 19 times? –  Henning Makholm Feb 4 '12 at 14:36
    
19 bacteria?? I'm not sure I understand exactly. –  David Feb 4 '12 at 14:42
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How do you manage to get $19$ by doubling anything? Were there $9.5$ bacteria five minutes earlier? –  Henning Makholm Feb 4 '12 at 14:43
    
No. We're just starting off with one and doubling every 5 min till 96 min. so you're right, it can't be 19. –  David Feb 4 '12 at 14:58
    
Okay. Doubling once gives 2. Doubling twice gives 4. Doubling three times gives 8. Doubling four times gives 16. Can you continue the sequence up to 19 times? –  Henning Makholm Feb 4 '12 at 15:00

2 Answers 2

up vote 4 down vote accepted

If you want to model the population with the formula: $$ P(t)=P_0 e^{kt}, $$ where $P_0$ is the initial population and $k$ is the exponential growth constant, then you must first find the values of $P_0$ and $k$.

We are told $P_0=1$, so $$ P(t)= e^{kt} $$

Let's now solve for $k$:

We know that after 5 minutes, the population is $2 $, so $$ 2 = e^{5k} $$ Solving the above for $k$ gives $k={\ln 2\over 5}$. So $$ P(t)=1\cdot e^{ {\ln 2\over 5}t} $$

After 96 minutes, the "population" is $$ P(96)=1\cdot e^{ {\ln 2\over 5}96}= e^{19.2 {\ln 2 } }\approx 602,248.763. $$

But, as @Henning Makholm points out in the comment below, this isn't realistic. The population at 96 minutes, assuming a bacteria splits in 2 every 5 minutes, is the population at $95$ minutes: $P(95)= e^{ {\ln 2\over 5}95}= e^{\ln 2\cdot19}=2^{19}$.


This could have been obtained more simply: 96 minutes is 19 doubling periods (plus an extra minute where no new bacteria are formed).

$\ \ \ $after 5 minutes, one doubling period, the population is 2.

$\ \ \ $after another 5 minutes, one more doubling period, the population is 4.

$\ \ \ $after another 5 minutes, one more doubling period, the population is 8.

$\ \ \ \ \ \ \ \ \vdots$

At 95 minutes, 19 doubling periods, the population will be $2^{19}$. At 96 minutes, the population will be $2^{19}$.

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it should be $2^{19}$ exactly, because the bacteria only divide after $5$ minutes. It's not as there will be one-and-a-half bacterium in the dish after three minutes. (If instead of a fixed 5-minute frequency we assume that each bacterium divides after random growth phase with an average length of 5 minutes, then the uncertainty in how long the first one takes to get going will make it impossible to give an answer to more than at most one significant digit anyway). –  Henning Makholm Feb 4 '12 at 14:52
    
@Henning Makholm Of course, thanks. –  David Mitra Feb 4 '12 at 15:05
    
@DavidMitra at the part where it says $e^{19.2 {\ln 2 } }$, where did you get 19.2 from? You didn't get it from dividing $\dfrac{ln2}{5}$ because $ln2$ is still there. Otherwise, this is exactly what I need! –  David Feb 4 '12 at 15:14
    
@David: 19.2 is the result of dividing 96 by 5. –  Henning Makholm Feb 4 '12 at 15:17

David Mitra's answer with Henning Makholm's input provides a good explanation and discussion of some technical details, but I'm going to disagree on two fronts.

First, using $P(t)=P_0 e^{kt}$ creates unnecessary work and makes the population model function needlessly complex (though I assume that it was used in that answer because it was suggested in the question). When we have an explicit description of growth/decay in terms of doubling (or other multiplying) time or half-life (or other fraction-time), we should use that explicitly in our model. If the initial population is $P_0$, then the population after $n$ doubling-times will be $P=P_0\cdot2^n$ because we multiply the population by $2$ for each of the $n$ doubling times. With a doubling time of 5 minutes, $t$ minutes is $\frac{t}{5}$ doubling times; starting with $P_0=1$, $$P(t)=2^{\frac{t}{5}}.$$ I will note that this is precisely equivalent to the formula from David Mitra's answer. I also dislike using $P(t)=P_0 e^{kt}$ to model this situation because there is commonly confusion about whether or not to use a sort of "continuous compounding" on top of the doubling-every-so-many-minutes.

Second, saying that the population doubles every 5 minutes is different from saying that each bacterium divides into two bacteria after existing for 5 minutes—I'm inclined to go with the interpretation that Henning Makholm discarded, that "each bacterium divides after random growth phase with an average length of 5 minutes," though I would think of it as having observed that the average population doubling time is 5 minutes. The problem is inherently one of modeling and as such there is some uncertainty in how to take the real-world phenomenon and map it to our mathematical model, then take the mathematical model result and map it back to the real world. This is particularly difficult in this instance because of starting with a single bacterium (which makes at least the first doubling and probably the first few doublings much more directly about single bacterium splitting rather than population doubling averages), but I would still stick to the simpler notion that what we have is a model based on observation and average behavior, so that we should go with the result from the fractional number of doubling times: $$P(96)=2^{\frac{96}{5}}\approx 602248.763.$$ This does leave an issue of how to interpret the fractional bacteria count returned by the model. Rounding to $602249$ is more consistent with the idea that this is an average-behavior model, but I would note that if this were a problem where there was a specific and precise discrete meaning (such as if the bacteria divided like clockwork, every 5 minutes, on the 5 minute mark, or if we were doing a financial problem about compound interest where interest was only paid at specific times), not having the whole of the 602249th bacterium would mean that our answer was $602248$. It can also be argued that giving the answer with the fractional bacteria count is the most correct, as it conveys the uncertainty in the model.

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