# bacteria problem

• May 5th 2010, 08:14 AM
ryan18
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.
• May 5th 2010, 08:59 AM
AllanCuz
Quote:

Originally Posted by 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 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 AM
Soroban
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: .$\displaystyle \ln(2^t) \:=\:\ln(10)\quad\Rightarrow\quad t\ln(2) \:=\:\ln(10)$

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