The base of the log can be accounted for by a constant scale factor, because, for example, if n is the number of bison,
log10(n)
= log10(e^ln(n))
= ln(n) log10(e)
and log10(e) is a constant.
This change of base is a linear scale on the logs.
Hence we can just take log 10 of the numbers of bison, and scale the answer by a constant factor which is log10(correct base), getting
7.778, 2.477 and 4.477
Scale that by about 2 = log10(100) to match the 5 bison in the middle pictogram, and there should be
16, 5, 9 bison on a logarithmic scale.
30,000 is roughly 1/3 of 60,000,000.
VERY roughly. Lol
It’s only off by roughly 20,000,000
Logarithmically scaled image. I’ll leave the determination of the base of the Log as an exercise for the viewer.
That’s what I thought, so I investigated.
The base of the log can be accounted for by a constant scale factor, because, for example, if n is the number of bison,
log10(n)
= log10(e^ln(n))
= ln(n) log10(e) and log10(e) is a constant.
This change of base is a linear scale on the logs.
Hence we can just take log 10 of the numbers of bison, and scale the answer by a constant factor which is log10(correct base), getting
7.778, 2.477 and 4.477
Scale that by about 2 = log10(100) to match the 5 bison in the middle pictogram, and there should be
16, 5, 9 bison on a logarithmic scale.
The diagram is also wrong if it’s logarithmic.
well, we know bison in the middle are worth approximately 75 each…
I would show my proof, but I don’t have enough space in this margin
I’m here for this comment all day.
Yeah, we need 799613 more bison images to justify the graphic.
yeah this graphic is terrible
I assumed someone would’ve already fixed it and went looking
…are you a bot trying to trick users into pedantically identifying images for your training data? Cus these are not what you claim the are.
hmm apparently they already know what those are, at least the last one
Anyway went “looking” and just thought it was funny the finding part I included without introduction featured those very wrong images :)