logaa = 1
loga1 = 0
loga (b • c) = logab + logac
loga (b / c) = logab - logac
logaxp = p • logax
logax = logbx / logba (формула перехода к новой основы)
logab = 1 / logba
logab = logapbp = p • logapb
alogab = b
logca • logab = logcb
logaα bβ = (β / α) • logab
alogcb = blogca
logaα b = logab / logaaα = (1 / α) • logab
logcalogcb = logcblogca
logcb • logca = logca • logcb
Внимание: log (a + b) ≠ loga + logb
Примечание: e = limn → ∞ (1 + 1 / n) n = e
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