bacteria problem

**ryan18** · May 5th 2010, 08:14 AM

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.

**AllanCuz** · May 5th 2010, 08:59 AM

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

$\frac{dm}{dt} = km$

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?

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

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

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

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

**Soroban** · May 5th 2010, 09:57 AM

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

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

Math Help - bacteria problem

LinkBack

Thread Tools

Display

bacteria problem

Similar Math Help Forum Discussions

bacteria problem

The no. of bacteria...

Did I do this bacteria growth problem correctly?

Bacteria Growth Problem

Bacteria Problem

Search Tags