well, suppose that x >= y, and that we are using the natural base: 
log(x + y) = log(x * (1 + y/x)) = log(x) + log(1 + y/x)

y/x = exp(log(y/x)) = exp(log(y) - log(x))

log(x + y) = log(x) + log(1 + exp(log(y) - log(x)))

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And that formula can be easily calculated using regular perl scalars because log(y) - log(x) is in [-log(x), 0] and exp(log(y) - log(x)) is in (0, 1].

In reply to Re: Add numbers in log domain by salva
in thread Add numbers in log domain [solved] by pRaNaV

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