Im trying to understand this write up [1] of cell population growth models and am confused about the use of natural logarithms. If cells double at a constant rate starting from 1 cell, then their cell number $N$ can be modeled as:
$N = 2^{x}$
where $x$ is the doubling step, or more simply, the units of time that have passed. To make $16$ cells, we need $\log_{2} = 4$ divisions or units of time to pass. Typically people use natural logs and $e$ to model this in the general form:
Growth equation: $N = e^{rt}$
where $r$ is the return on the number of cells and $t$ is the time that has passed.
My questions are:
(Q1) Why is $e$ a relevant base here in place of $2$, assuming that cells double and do not divide into portions? If we want to know how much time needs to pass to get 16 cells and use this equation then:
$16 = e^{rt}$
Assume $r = 1$ since each cell doubles, then:
$\ln(16) = t\ln(e)$
so $t = \ln(16) = 2.7$, which is counterintuitive. We can substitute $e^{\ln(2)}$ in place of $e$ to get the right number, but my question is why do we bother with $e$ at all? Can someone explain the difference between saying that it takes $\ln(16)$ units in base $e$ to divide versus $\log_{2}(16)$ units in base $2$? Is it fair to say that these statements are identical because the logarithms scale similarly?
The paper [1] says that they have the same curvature, but not sure what that means in this case.
(Q2) know that $e$ is useful when you have compounded returns or when you look at continuous time, as in interest on an investment. But there are no compounded returns here as far as I can tell since you cannot get fractions of cells made though division. Similarly, the $e$ growth curve is continuous but the growth is discrete. Am I missing something? Are these relevant here?
(Q3) I can see the argument for using something other than $N = 2^{x}$ since cells do not grow in staircase fashion predicted by $2^{x}$ since they are not normally synchronized. Does the $e$-based equation capture this better than the $2^{x}$ equation? It seems to me that not, that the $e$-based equation is just continuous but does not do anything special to account for independent cells that divide according to their own slightly different rates (e.g. each following a binomial model of division where each cell gets a slightly different rate of division.)
Any clarification on the use of this model would be appreciated. I understand the relevance to finance and other continuous domains where you have compounded returns on a financial investment, but I am hard pressed to find an argument for using $e$ over $2$ for population growth of dividing cells.
[1] http://www.mpi-bremen.de/Binaries/Binary13037/Wachstumsversuch.pdf
Best Answer
if you are using $e^{rt}$, then since doubling time is (in your chosen unit of time) equal to $1$, we get $e^{(r)(1)}=2$. Taking logarithms to the base $e$, we get $r=\ln 2$.
It simplifies things to use the doubling time as the unit of time. However, in a more complicated situation, where we have say two kinds of cell, what unit shall we use? In the long run, the conventional units of time are more useful.
So the population at time $x$ is $e^{(\ln 2)(x)}$. Since $e^{\ln 2}=2$, this is just a fancy way of writing $2^x$.
Both models can be considered as continuous models, since $x$ is not restricted to be an integer. They are in fact the same model. And yes, we do get results that are not integers, but that is not important. No mathematical model will represent a complex biological phenomenon exactly. If the numbers involved are large, an "error" of half a cell has no practical significance. Indeed, the error is likely to be far larger than that.
The function $e^t$ has nice technical properties. For example, its derivative is $e^t$. the derivative of $2^t$ is the more messy $(\ln 2)2^t$. But anything one can do by using base $e$ can also be done using base $2$, or $10$. There will be small differences of detail, that's all.