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 2 hours. Jaquan has a different type of bacteria that doubles every 3 hours. How many bacteria should Jaquan start with so that they have the same amount at the end of the day?

Answer 1

800

Explanation:

Answer 2

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