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

Here's the general formula for bacteria growth/decay problems

Af = Ai (e^kt)

where:
Af = Final amount
Ai = Initial amount
k = growth rate constant
t = time

But there's another formula for a doubling problem.
kt = ln(2)

So, Colby (1)
k1A = ln(2) / t
k1A = ln(2) / 2 = 0.34657 per hour.

So, Jaquan (2)
k2A = ln(2) / t
k2A = ln(2) /3 = 0.23105 per hour.

We need to use the rate of Colby and Jaquan in order to get the final amount in 1 day or 24 hours.

Af1 = 50(e^0.34657(24))
Af1 = 204,800

Af2 = 204,800 = Ai2(e^0.23105(24))
Af2 = 800