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Pre-Calc (exponential growth)

A COLONY OF BACTERIA IS GROWN UNDER IDEAL CONDITIONS IN A LAB SO THAT THE POPULATION INCREASES EXPONENTIALLY WITH TIME. At the end of the three hours, there are 10,000 bacteria. At the end of the 5 hours, there are 40,000 bacteria. How many bacteria were present initially?

I was planning to use Y=P(O)e^kt
but I'm not sure how to set it up.

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Phillip

  1. Y(5)=40,000=Pe^5k
    y(3)=10,000=Pe^3k

    divide one equation by the other

    4=e^2k solve for k. Put it back in, and solve for P(O)

    There is an easier way. From the data, one can see that population doubles each hour.

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    bobpursley

  2. Sure, use y = Po e^kt
    when t = 3:
    10,000 = Po e^3k
    ln 10,000 = ln Po + 3k ln e but ln e is 1
    so
    ln 10,000 = ln Po + 3 k
    similarly
    ln 40,000 = ln Po + 5 k

    subtract
    ln 10,000 -ln 40,000 = -2 k

    ln (10,000/40,000) = -2k
    -1.386 = -2k
    k = .693

    then back
    10,000 = Po e^(2.0794)
    Po = 1250

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    Damon

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