A bacteria population is initially 320. After 45 minutes, they've grown in number to 700. What is the doubling time for this population? ?

39 minutes 41 seconds

39 minutes 31 seconds

39 minutes 21 seconds

39 minutes 51 seconds

3 Answers

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  • 4 months ago

    A bacteria population is initially 320. After 45 minutes, 

    they've grown in number to 700. 

    What is the doubling time for this population?

    Bacteria population's doubling time for this population is 

    39 minutes 51 seconds

    Explanation:

    Let the doubling time of bacteria population be t.  

    This means that after t minutes, it will be doubled and after nt minutes

    it will grow 2^n  times. 

    Hence in the instant case

    700 /320 = 2^(45/t)   or  21.875 = 2^(45/t) 

    or log 2 ( 2.1875 ) = 45/t

    or log 2.1875 log 2 = 45/t

    or *0.33995/0.30103  = 45/t # 

    i.e. t = 45⋅0.30103

    0.33995 = 39.05  minutes

    Hence bacteria population's doubling time for this population is

    39.05  minutes

    39 minutes 51 seconds

  • 4 months ago

    Assuming exponential growth:

    y = ab^x

    a = initial population = 320

    y = 320 * b^x

    When x = 45, y = 700:

    700 = 320 * b^45

    b^45 = 700/320 = 35/16 = 2.1875

    b = 2.1875^(1/45)

    y = 320 * 2.1875^(x/45)

    After the doubling time, y is 2 * 320 = 640:

    320 * 2.1875^(x/45) = 640

    2.1875^(x/45) = 640 /320 = 2

    x/45 = log[2.1875](2)

    x = 45*log[2.1875](2) =~ 39.848292522575846712897275462287

    Rounding off, that's 39 minutes and 51 seconds.

  • 4 months ago

    B(t) = 320 x e^kt

    When t = 45, B(t) = 700

    so, 700 = 320 x e^45k

    => e^45k = 35/16

    so, 45k = ln(35/16)

    i.e. k = (1/45)ln(35/16)

    Then, B(t) = 320 x e^[(t/45)ln(35/16)]

    Hence, B(t) = 320 x (35/16)^(t/45)

    Doubling occurs when B(t) = 640

    i.e. when (35/16)^(t/45) = 2

    so, (t/45)ln(35/16) = ln2

    Hence, t = 45ln2/ln(35/16) => 39 minutes and 51 seconds

    :)>

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