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17 December, 20:55

Suppose that the concentration of a bacteria sample is 60 comma 000 bacteria per milliliter. If the concentration triples in 4 days, how long will it take for the concentration to reach 102 comma 000 bacteria per milliliter?

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  1. 17 December, 21:25
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    1.932 days (or approximatelly 1 day, 22 hours and 22 minutes)

    Step-by-step explanation:

    The inicial concentration is 60,000, and this concentration triples every 4 days, so we can write the equation:

    P = Po * r^t

    where P is the final concentration after t periods of 4 days, Po is the inicial concentration and r is the ratio that the concentration increases (r = 3)

    Then, we have that:

    102000 = 60000 * 3^t

    3^t = 102/60 = 1.7

    log (3^t) = log (1.7)

    t*log (3) = log (1.7)

    t = log (1.7) / log (3) = 0.483

    so the number of days that will take is 4*0.483 = 1.932 days (or approximatelly 1 day, 22 hours and 22 minutes)
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