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?
[Math] Exponential growth function
algebra-precalculusexponential functionfunctions
Best Answer
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.