Luria-delbruck experiment question

[daniel6874]

My question is about the classic study cited below (*) and available as a free pdf download from

http://www.genetics.org/content/28/6/491.full.pdf+html

At page 495 the authors use the "average division time" of bacteria dt, divided by ln(2), as the time unit in the equations

(1) dN_t /dt = N_t , (2) N_t = N_o e^t.

Can anyone tell me why this division by ln(2) was done? N_o is the original number of bacteria present. N_t is the number at time t.

Obviously it has something to do with integration/differentiation, but I am missing the point. Thanks.



*Luria, S. E.; Delbrück, M. (1943). "Mutations of Bacteria from Virus Sensitivity to Virus Resistance". Genetics 28 (6): 491–511.

 

[Ygggdrasil]

The doubling time divided by ln(2) gives the appropriate time constant for an exponential growth. Writing the equation for exponential growth in terms of the doubling time (t_d) gives:

N(t) = N_o 2^(t/t_d)

Equivalently, this can be written as:

N(t) = N_o e^(t ln(2) / t_d) = N_o e^(t/t_c)

Where t_c = t_d/ln(2). This makes use of the fact that e^ln(2) = 2.

 

[bobze]

Bacteria double exponentially; 2 become 4, 4 become 8, etc. This describes Log in base 2 (2², 2³, etc).

The application then is used to describe things that grow or decay exponentially. A more detailed explanation can be found http://logbase2.blogspot.com/2007/12/log-base-2.html" [Broken]

 

[daniel6874]

...
N(t) = N_o 2^(t/t_d)

Equivalently, this can be written as:

N(t) = N_o e^(t ln(2) / t_d) ...

This is crystal clear. I forgot he was starting with 2^t/t_d as a growth law. Thanks!