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.
Best Answer
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 $2^{19}$.