# How many bacteria would be present 15 hours after the experiment began if a set of bacteria begins with 20 and doubles every 2 hours?

Dec 7, 2014

The answer is $3811$ bacteria.

Using a function to describe exponential growth, we can say that

$A = {A}_{\circ} \cdot {e}^{k \cdot t}$, where

$A$ - is the amount we need to find out (in this case, the number of bacteria after 15 hours of growth);
${A}_{\circ}$ - the initial number of bacteria;
$k$ - growth rate;
$t$ - time;

We were given $t$=$15$ hours and ${A}_{\circ}$=$20$ bacteria; however, both $k$ and $A$ need to be determined.

We will determine $k$ by using the fact that the number of bacteria doubles every two hours - this means that after the first 2 hours, we will have 40 bacteria. So,

$40 = 20 \cdot {e}^{k \cdot 2}$, which gives us a $k$ = $0.35$.

Therefore, $A = 20 \cdot {e}^{0.35 \cdot 15} = 3811$ bacteria.