The way people normally plot these kinds of distributions is on a log/log graph (to be clear: where both axes are log'd, not where only one of the axes is doubly-log'd).
That's not true, only if the distribution in both directions is exponential, which is not the case here. The horizontal scale is just the rank, which is linear.
Interesting; FWIW, I am going based on the comment "the channels follow a power law" from the grandparent post. I was under the (possibly mistaken) impression that these are normally plotted using log-log; would you be willing to explain what I am getting wrong? (I am totally happy to believe I'm missing something btw, and I certainly did not anticipate the person I responded to deleting their post...).
(edit: Further, looking at the graphs, it certainly seems like they would do well on a log-log chart; also, I would have expected the behavior that was described in the deleted post from a distribution of this sort: specifically, where further introductions of logarithms to the same axis did not affect the result, as they were asymptotically identical.)
The parent is obviously wrong about power law (like e^x). It's not even e^(e^x), it's something that grows more rapidly, like x^x, or x factorial. I'm too lazy to figure it out.
Rank can be considered linear or logarithmic. It can be argued that the difference between being 1st or 10th is orders of magnitude bigger than the difference between being 100th and 110th.
In fact I think a log scale for rank usually makes the most sense, especially when starting from 1st.
The parent is correct, the claimed relationship being y ~ rank^a for some a<0. You would want to see a linear relationship on the log log plot. Another choice would be to examine the complmentary CDF P(X > x) and look for power law tails.
I don't really have time to do this because I'm supposed to be working on a paper about, wait for it, power laws, but I went ahead and re-plotted the data using log-log axes. Sure enough, it's linear as would be expected for power law type behavior: