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ab = c
b = logac
loga1 =
0
logaa =
1
logca + logcb =
logc(ab)
logca - logcb =
logc(a/b)
logcan =
nlogca
logcb / logca =
logab
logeA is written as
lnA
y = ex →
x = ln y
ln(e) =
1