Anonymous
Anonymous asked in Science & MathematicsMathematics · 1 month ago

Math Homework?

The count in a bateria culture was 300 after 15 minutes and 1700 after 40 minutes. Assuming the count grows exponentially,

What was the initial size of the culture?

Find the doubling period.

Find the population after 85 minutes.

When will the population reach 14000. 

1 Answer

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  • P(15) = 300

    P(40) = 1700

    P(t) = a * 2^(t/k)

    k is the doubling time

    a is the initial population

    a * 2^(40/k) = 1700

    a * 2^(15/k) = 300

    a * 2^(40/k) / (a * 2^(15/k)) = 1700/300

    2^(40/k - 15/k) = 17/3

    2^(25/k) = 17/3

    2^(1/k) = (17/3)^(1/25)

    P(t) = a * 2^(t/k)

    P(t) = a * (2^(1/k))^t

    P(t) = a * ((17/3)^(1/25))^t

    P(t) = a * (17/3)^(t/25)

    Isn't that lovely?  You can get the same effect by starting out with e, or 10, or a googol, or whatever.  It doesn't have to be a 2, but I felt that by using a 2, you could see why the population doubles every k minutes.  But it has served its purpose and now we have something much nicer to look at.  Now, let's solve for a.

    P(15) = 300

    300 = a * (17/3)^(15/25)

    300 = a * (17/3)^(3/5)

    300 / (17/3)^(3/5) = a

    a = 300 * (17/3)^(-3/5)

    a = ‭105.9557804410011235032485352032‬...

    Round to the nearest whole value.

    a = 106

    There were 106 bacteria initially.

    2a = a * (17/3)^(t/25)

    2 = (17/3)^(t/25)

    2^25 = (17/3)^t

    25 * ln(2) = t * ln(17/3)

    25 * ln(2) / ln(17/3) = t

    t = ‭9.990008630613594640745895492379‬...

    9.99 minutes is the doubling time

    P(85) = 106 * (17/3)^(85/25)

    P(85) = 106 * (17/3)^(17/5)

    P(85) = 106 * (17/3)^(3.4)

    P(85) = ‭38,602.900010311218626673704375427‬....

    Or more accurately

    P(85) = 300 * (17/3)^(-3/5) * (17/3)^(17/5)

    P(85) = 300 * (17/3)^(14/5)

    P(85) = 300 * (17/3)^(2.8)

    P(85) = ‭38,586.796206400523465759934256926‬

    Take your pick, round accordingly.  I was always taught to save all of my rounding until the very end, but that's just me.

    14000 = 106 * (17/3)^(t/25)

    14000/106 = (17/3)^(t/25)

    (14000/106)^(25) = (17/3)^t

    25 * ln(14000/106) = t * ln(17/3)

    t = 25 * ln(14000/106) / ln(17/3)

    Or....

    14000 = 300 * (17/3)^(-3/5) * (17/3)^(t/25)

    140/3 = (17/3)^(t/25 - 15/25)

    (140/3) = (17/3)^((t - 15) / 25)

    (140/3)^(25) = (17/3)^(t - 15)

    25 * ln(140/3) = (t - 15) * ln(17/3)

    25 * ln(140/3) + 15 * ln(17/3) = ln(17/3) * t

    t = (25 * ln(140/3) + 15 * ln(17/3)) / ln(17/3)

    Wolfram Alpha can evaluate those for you, if you copy and paste them exactly.

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