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  1. #1
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    bacteria problem

    At the beginning of an experiment a colony has N bacteria. Two hours later it has 4N bacteria. How many hours, measured from the beginning, does it take for the colony to have 10N bacteria.
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  2. #2
    Senior Member AllanCuz's Avatar
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    Quote Originally Posted by ryan18 View Post
    At the beginning of an experiment a colony has N bacteria. Two hours later it has 4N bacteria. How many hours, measured from the beginning, does it take for the colony to have 10N bacteria.
    The equation for bacteria growth comes from

     \frac{dm}{dt} = km

    Which is a well known seperable equation. In this case, I imagine you don't need to derive the equation so we'll skip right to it.

    We are going to use

     N_t = N e^{kt} where we let t be time in hours.

    At time 0 we have N bacteria so the above equation stays the same.
    At time 2 we have 4N bacteria so we can find K,

     4N = N e^{k2}

     4=  e^{k2}

     ln4=  k2

     \frac{ln4}{2}=  k

    Now, we sub back into our first equation to get.

     N_t = N e^{\frac{ln4}{2} t}

    At what time t does the colonie have 10 bacteria?

     10N= N e^{\frac{ln4}{2} t}

     10=  e^{\frac{ln4}{2} t}

     ln10=  \frac{ln4}{2} t

     \frac{2ln10}{ln4} =  t
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  3. #3
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    Hello, ryan18!

    At the beginning of an experiment a colony has N bacteria.
    Two hours later, it has 4N bacteria.
    How many hours, measured from the beginning,
    . . does it take for the colony to have 10N bacteria?

    The general population function is: . P \;=\;P_oe^{kt}


    When t = 0,\;P = N\!:\;\;N \:=\:P_oe^0 \quad\Rightarrow\quad P_o \,=\,N

    . . We have: . P \;=\;Ne^{kt}


    When t = 2,\;P = 4N\!:\;\; 4N \:=\:Ne^{2k} \quad\Rightarrow\quad e^{2k} \:=\:4

    \text{Take logs: }\;\ln(e^{2k}) \:=\;\ln(4) \quad\Rightarrow\quad 2k\cdot\underbrace{\ln(e)}_{\text{This is 1}} \:=\:\ln(4)  \quad\Rightarrow\quad 2k \:=\:\ln(4)
    . . k \;=\;\tfrac{1}{2}\ln(4) \;=\; \ln\left(4^{\frac{1}{2}}\right) \;=\;\ln(2)


    The function is: . P \;=\;Ne^{\ln(2)\,t} \;=\;N\left(e^{\ln(2)}\right)^t \quad\Rightarrow\quad P \;=\;N\cdot 2^t


    If P = 10N, we have: . N\cdot2^t \:=\:10N \quad\Rightarrow\quad 2^t \:=\:10

    Take logs: . \ln(2^t) \:=\:\ln(10)\quad\Rightarrow\quad t\ln(2) \:=\:\ln(10)

    . . t \:=\:\frac{\ln(10)}{\ln(2)} \;=\;3.321928095\text{ hrs} \;\approx\;3\text{ hr, }19\text{ min, }19\text{ sec}

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