Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Under ideal conditions a certain bacteria population is know to double every three hours. Suppose that there are initially 100 bacteria. How would you go about formulating a function for this?

share|cite|improve this question
up vote 0 down vote accepted

The general approach described by Todd Wilcox is the best way to go about things: It is important to become thoroughly familiar with the function $e^{rt}$.

When doubling time is given, there is a quick formula that works. Let $d$ be the doubling time. Then if $P(t)$ is the population at time $t$, we have $$P(t)=P(0)\,2^{t/d}.$$ So in our case, $$P(t)=(100)2^{t/3}.\tag{$1$}$$ The general approach, which you should carry out, will give you after some manipulation $$P(t)=(100)e^{t \log 2/3},$$ where by $\log$ we mean logarithm to the base $e$ ($\ln$ on your calculator). This is the same as the answer $(1)$, since $$e^{x\log 2}=\left(e^{\log 2}\right)^x=2^x,$$ since $e^{\log 2}=2$.

Remark: As long as one is willing to believe that the answer has shape $P(t)=A\,2^{kt}$, the formula is easy to prove. For put $t=0$. Then $P(0)=A\,2^{(k)(0)}=A$, so $A=P(0)$. Also, since doubling time is $d$, we have $P(d)=2P(0)$. But $P(t)=P(0)\,2^{kt}$. Put $t=d$. We get $2P(0)=P(0)\,2^{kd}$, and therefore $kd=1$, meaning that $k=1/d$. This yields the formula $P(t)=P(0)\,2^{t/d}$ mentioned above.

share|cite|improve this answer

All exponential growth or decay follow the same basic mathematical relationship:$$y(t)=Pe^{rt},$$where $y$ represents the amount of whatever is growing or decaying at time $t$, $P$ is the "principle" (taken from compounding interest on money) or the starting value, and $r$ is the rate of growth or decay.

Often, as in your question, we have to find $r$ using other known quantites. In your case, we know that if we start with $P$ bacteria, we should have $y(3)=2P$, or double the initial amount after three hours. So we can set up and then solve for $r$:$$2P=Pe^{3r}.$$

Once you have $r$ and the initial amount (100 for the second half of your problem), you can find the final amount at any time $t$. Does that help?

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.