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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- May 5th 2010, 08:14 AMryan18bacteria 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.

- May 5th 2010, 08:59 AMAllanCuz
The equation for bacteria growth comes from

$\displaystyle \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

$\displaystyle 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,

$\displaystyle 4N = N e^{k2} $

$\displaystyle 4= e^{k2} $

$\displaystyle ln4= k2 $

$\displaystyle \frac{ln4}{2}= k $

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

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

At what time t does the colonie have 10 bacteria?

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

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

$\displaystyle ln10= \frac{ln4}{2} t $

$\displaystyle \frac{2ln10}{ln4} = t $ - May 5th 2010, 09:57 AMSoroban
Hello, ryan18!

Quote:

At the beginning of an experiment a colony has $\displaystyle N$ bacteria.

Two hours later, it has $\displaystyle 4N$ bacteria.

How many hours, measured from the beginning,

. . does it take for the colony to have $\displaystyle 10N$ bacteria?

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

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

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

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

$\displaystyle \text{Take logs: }\;\ln(e^{2k}) \:=\;\ln(4) \quad\Rightarrow\quad 2k\cdot\underbrace{\ln(e)}_{\text{This is 1}} \:=\:\ln(4) \quad\Rightarrow\quad 2k \:=\:\ln(4)$

. . $\displaystyle k \;=\;\tfrac{1}{2}\ln(4) \;=\; \ln\left(4^{\frac{1}{2}}\right) \;=\;\ln(2) $

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

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

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

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