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?

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