I always get a little annoyed when engineers or math textbooks use the word "antilogarithm." Isn't it just exponentiation? Like if $\log(2) \approx 0.301$, then $10^{0.301}\approx 2$ . Why say "antilogarithm?" Is there some subtly different meaning? Am I missing something?

  • 11
    $\begingroup$ As far as I know, there isn't a different meaning. I guess it's just to utilize the prefix "anti-" to emphasize that it's an inverse operation. Sort of like how differentiation has antidifferentiation. $\endgroup$ Commented Aug 10, 2020 at 3:51
  • 2
    $\begingroup$ The only disadvantage is the length of the terminology or saying it out loud. "The antilogarithm of base e" $\endgroup$
    – David P
    Commented Aug 10, 2020 at 3:57
  • 7
    $\begingroup$ In the old days, when people used logarithm tables to do multiplication, the page after the log tables was the antilog tables (to base ten of course). You looked up the logs of your numbers, added them, and then looked up the antilog of that. $\endgroup$ Commented Aug 10, 2020 at 5:36
  • 2
    $\begingroup$ @AnginaSeng Odd. There were no exponentiation (or "antilogarithm") tables in the logarithm tables I used. After adding/subtracting the logarithms, one looked up the mantissa in the log table (easy, since the logarithm is montonic), getting $10^m$ from the margins (manual interpolation if needed). $\endgroup$ Commented Aug 10, 2020 at 8:59
  • 6
    $\begingroup$ Have you read the prolog? $\endgroup$ Commented Aug 10, 2020 at 13:11

1 Answer 1



First, when Napier invented logarithms, his application was not the inverse of the exponential function. He was originally multiplying quantities called "sines", which are not what you're thinking when you see that word. These "sines" are vaguely similar to the positions of the arrow in Zeno's paradox of the arrow. To each "sine" (a quantity) was associated a quantity, the logarithm of that "sine". Napier arranged for his logarithms to have the property that when the "sines" decreased in geometric proportion, the logarithms increased in arithmetic proportion. So he transformed multiplication of "sines" into addition of their logarithms.

Now to your question: to go from a "sine" to its logarithm was performed by table lookup. To go from a logarithm to its "sine" was performed by backwards table lookup. So, quite literally, an antilogarithm is found by using the table backwards. Since we are not working on usual quantities, but on "sines", the inverse operation is not exponentiation.

Napier and Briggs then modified the logarithm to work with "normal" quantities instead of "sines". At this point, the inverse of the logarithm was exponentiation.

Note that the inverse of the logarithm couldn't be called exponentiation by Napier since he was writing in 1614 and subsequently. In 1748 Euler wrote "consider exponentials or powers in which the exponent itself is a variable" ("Primum ergo considerandæ sunt quantitates exponentiales, seu Potestates, quarum Exponens ipse est quantitas variabilis.", from Introductio in analysin infinitorum) , which seems to be the first time the exponent was not a constant positive integer. Until we make the generalization of exponents to arbitrary powers, there is no hope of describing the inverse logarithm as an exponential function.

One "convenience" of the antilog notation is that the following equation $$ \log \antilog x = x = \antilog \log x $$ is true both for Napier's "sines" and subsequent inverse exponential logarithms. Rewriting this where the base is variable (which is not what Napier was considering) $$ \log_b \antilog_b x = x = \antilog_b \log_b x \text{,} $$ which is (as many students have shown) more parsable than $$ \log_b b^x = x = b^{\log_b x} \text{.} $$

  • 4
    $\begingroup$ Wow. Thank you for the research. $\endgroup$
    – Palbitt
    Commented Aug 10, 2020 at 5:45
  • 3
    $\begingroup$ Very interesting answer. $\endgroup$ Commented Aug 10, 2020 at 10:56
  • 3
    $\begingroup$ Interesting; I was expecting this use of "sines" to then some how be related to the fact that e^(sqrt(-1)x = cos(x) + sqrt(-1) sin(x) $\endgroup$
    – Foon
    Commented Aug 10, 2020 at 14:54
  • 14
    $\begingroup$ @Foon : It's not; that has to wait 100 years for Euler. I understood what "sines" were several years ago, but that understanding has atrophied away. I recall that the idea is based on the product to sum formulas to convert the product of (usual) sines and cosines to a sum of sines and cosines. This was how Napier captured the logarithmic property of products turning into sums. (Also, he was doing computational astronomy, so trig identities were very familiar...) $\endgroup$ Commented Aug 10, 2020 at 16:50
  • 2
    $\begingroup$ Great answer (+1). I think it was Wallis around 1650 and then Newton (with his general binomial theorem) who first considered arbitrary exponents, although hints of these go back to Oresme. $\endgroup$ Commented Aug 10, 2020 at 16:57

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .