Multiplying powers means adding exponents, so with a log and antilog table in hand multiplication becomes one addition and two lookups. Today you write the fast multiply and confirm it agrees with the slow one from lesson 3 - the payoff that closes the field chapter.
Multiply via log and antilog tables, matching the slow multiply and handling zero.
Here is the reward for building both tables. Multiplying two powers of the same base means adding their exponents: 2^m times 2^n is 2^(m+n). So to multiply a by b, take log[a] and log[b], add them, and look the sum back up in exp. Because the powers cycle with period 255, you reduce the exponent sum modulo 255 (or index a doubled 512-entry table and skip the modulo). One addition, two lookups, done - no per-bit loop.
The only trap is 0: it has no logarithm, so a zero operand must short-circuit to 0 before you touch the log table. Check gmul(2, 128): log[2]=1, log[128]=7, sum 8, and exp[8]=29 - the very reduce case you traced by hand in lesson 2, now falling out of a lookup. Confirm the fast multiply equals your slow gmul on the same inputs and you can retire the slow one from the hot path. This little routine is the workhorse of the entire Reed-Solomon chapter that starts next.
// a*b = exp[log a + log b]. Add exponents mod 255. Either// operand being 0 short-circuits to 0 (0 has no logarithm).func gmulFast(a, b byte) byte {if a == 0 || b == 0 {return 0}return exp[(int(logt[a])+int(logt[b]))%255]}