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I agree that logarithms are numbers, and Napier chose arithmós reasonably. But why did Napier pick Greek logós ‘ratio’? I don't see a ratio in this definition of the logarithm in James Stewart, Calculus 2011 7e (not the Early Transcendentals version), p 404.

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Word Origins (2005 2e) by John Ayto. p 314 Right column.

logarithm [17]

Greek lógos had a remarkably wide spread of meanings, ranging from ‘speech, saying’ to ‘reason, reckoning, calculation’, and ‘ratio’. The more ‘verbal’ end of its spectrum has given English the suffixes -logue and -logy (as in dialogue, tautology, etc), while the ‘reasoning’ component has contributed logic [14] (from the Greek derivative logiké), logistic [17] (from the Greek derivative logistikós ‘of calculation’), and logarithm, coined in the early 17th century by the English mathematician John Napier from Greek logós ‘ratio’ and arithmós ‘number’ (source of English arithmetic [13]).

I don't know if zbeckabee's quotation here matches Napier's rationale.

It comes from the Greek words 'logos' and 'arithmes.' The second word means 'number'. The first word has a broad range of meanings, starting from 'word' or 'speech' (thus our word 'dialog'), going into 'thought' and 'reason' (thus 'logical'), and on to 'proportion'. In this range of meanings, it resembles the Latin 'ratio', which means both 'reason' and 'proportion' (thus the two meanings of 'rational': reasonable, and expressable as a ratio or proportion of two integers).

Thus, a 'logarithm' is a 'proportion number'. I suppose the thinking was that logarithms help in the calculation of proportions. Instead of multiplying a number by a fraction, you just add or subtract the logarithm of the fraction to the logarithm of the number.

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    $\begingroup$ If you can read Latin, look at the original publication: John Napier, "Mirifici Logarithmorum Canonis Descriptio", Edinburgh: Andrew Hart 1614. My Latin is very rusty, but in the first chapter Napier shows (p. 4) two lines next to each other: one is marked at equidistant points, the other such that there is a constant ratio in the distance between neighboring points. Logarithms provide the mapping between these lines. $\endgroup$
    – njuffa
    Mar 18, 2021 at 9:08
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    $\begingroup$ If I understand Napier correctly, he states that the logarithm provides a count of how often a ratio has been applied. $\endgroup$
    – njuffa
    Mar 18, 2021 at 9:36
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    $\begingroup$ See Logarithm $\endgroup$ Mar 18, 2021 at 13:17
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    $\begingroup$ Napier does not explicitly state why he chose the abbreviation of logos arithmos. As was pointed out by Hacker, it involves a wordplay. One possible translation is "reckoning numbers" (see e.g. Caulfield), the other refers to Napier's model that converted logoi (ratios) to arithmoi (numbers), as njuffa described, more precisely, progressions in constant ratio to progressions in constant increment (unit), which is how Greeks understood "numbers". $\endgroup$
    – Conifold
    Mar 18, 2021 at 16:09
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    $\begingroup$ There was no notion of (real) number at the time of Napier. "Numbers" were only integers. A modern calculus book is a poor guide to the work of 16 century. $\endgroup$ Mar 19, 2021 at 12:46

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Your specific question is

I don't see a ratio in this definition of the logarithm in James Stewart, [...]

The ratio is implicit in that definition, but it is explicit in the change of base formula $$ \log_a(x) = \ln(x)/\ln(a) $$ which equates a logarithm to a given base to the quotient (or ratio) of natural logarithms. In fact, logarithms to a given base are exactly similar to measurement of a quantity in a given unit of measure. That is, they are the ratio of a given quantity relative to the unit quantity. That is how you can conceptualize logarithms as the relative ratio of the growth rate of two geometric progressions. One based on the given growth rate and the other based on the standard unit growth rate. For example, the geometric progression $\,1,a^2,a^4,a^6,\dots\,$ grows at twice the rate of the geometric progression $\,1,a,a^2,a^3,\dots\,$ in the sense that the first takes $\,1/2\,$ as many steps to grow to a given size than the other. This interpretation agrees with the result $\,\log_a(a^2) = 2.\,$

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