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28 October, 01:11

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?

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  1. 28 October, 01:26
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    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
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