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I'm learning about finite fields and I came across this example online: http://www.csee.umbc.edu/~lomonaco/s12/443/handouts/Log-Antilog-Calculation.pdf

I'm having trouble understanding what exactly log/antilog tables are used for, and how the calculations are being done in this example. I'm guessing since it's GF(2^4) = GF(16) so there's 16 entries and the highest exponent is 4 which means I have to use at most E^4 in the calculations (I can't get that symbol correctly displayed here, using E for now).

So what exactly are the uses for these tables and how do the calculations work?

Edit: link didn't work, example pic below.

enter image description here

pfinferno
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    Your link did not work for me. I don't know if it helps you, but I prepared similar tables for $GF(8)$ and $GF(16)$ here together with some examples - probably not enough. If you want something more specific to be added there, just ask. – Jyrki Lahtonen Oct 12 '15 at 05:59
  • Thanks for letting me know, edited the post to include a picture of the problem. – pfinferno Oct 12 '15 at 15:43

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How to use a log table? Oldsters like me know that if you want to multiply two quantities, like $\xi^3+\xi^2$ and $\xi^4+\xi^3+\xi^2+\xi$ in $\Bbb F_{16}$, you look up their logarithms, to find $6$ and $13$. Then add the logarithms modulo $15$ to get $4$. The quantity whose log is $4$ is $\xi+1$, and that’s your product.

To divide, subtract the logarithms, modulo $15$.

Lubin
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  • How are you getting 4 though? If you add the two quantities of 6 and 13 together you get ξ^4 + ξ – pfinferno Oct 12 '15 at 20:03
  • I get ξ^4 = ξ + 1, so it would be ξ + ξ + 1 = 1. But why is 4 = ξ^4 + 1? – pfinferno Oct 12 '15 at 20:37
  • Nevermind, I was completely forgetting that ξ^4 = ξ + 1 because ξ^4 + ξ + 1 = 0. Since subtraction and addition are the same over GF(2) it makes ξ^4 = ξ + 1. – pfinferno Oct 12 '15 at 22:35
  • $6+13=19\equiv4\pmod{15}$. You add the logarithms, not their entries. – Lubin Oct 13 '15 at 02:47