Question

colby and jaquan are growing bacteria in an experiment in a laboratory. Colby starts with 50 bacteria in his culture and the number of bacteria doubles every two hours. Jaquan has a different type of bacteria that doubles every three hours. How many bacteria should Jaquan start with so that they have they have the same amount at the end of the day?

Answer 1

To get started, we will use the general formula for bacteria growth/decay problems:

$A_{f} = A_{i} ( e^{kt} )$

where: 
A_{f} = Final amount
A_{i} = Initial amount
k = growth rate constant
t = time


For doubling problems, the general formula can be shortened to:

$kt = ln(2)$

Now, we can use the shortened formula to calculate the growth rate constant of both bacteria:

Colby (1):
$k_{1} = ln(2)/t$
$k_{1} = ln(2)/2 = 0.34657$ per hour

Jaquan (2):
$k_{2} = ln(2)/t$
$k_{2} = ln(2)/3 = 0.23105$ per hour

Using Colby's rate constant, we can use the general formula to calculate for Colby's final amount after 1 day (24 hours).

Note: All units must be constant, so convert day to hours.

$A_{f1} = 50( e^{0.34657(24)})$
$A_{f1} = 204,800$

Remember that the final amount for both bacteria must be the same after 24 hours. Again, using the general formula, we can calculate the initial amount of bacteria that Jaquan needs:

$A_{f2} = 204,800 = A_{i2} ( e^{0.23105(24)} )$
$A_{i2} = 800$