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.
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.
I would show my proof, but I don’t have enough space in this margin
I’m here for this comment all day.
well, we know bison in the middle are worth approximately 75 each…
OBVIOUSLY!!