# Luria-delbruck experiment question

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.

## Answers and Replies

Ygggdrasil
Science Advisor
Gold Member
2019 Award
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 (td) gives:

N(t) = No 2^(t/td)

Equivalently, this can be written as:

N(t) = No e^(t ln(2) / td) = No e^(t/tc)

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

bobze
Science Advisor
Gold Member
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]

Last edited by a moderator:
...
N(t) = No 2^(t/td)

Equivalently, this can be written as:

N(t) = No e^(t ln(2) / td) ...

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