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A Taylor series expansion of ln(1-p) gives...
ln(1-p) = - p - p*p/2 - p*p*p/3 - p*p*p*p/4 - ...
Thus, replacing ln(1-p) with p is only valid when
p/2 + p*p/3 + p*p*p/4 + ... << 1
... one could then ask why we use T = ln(2) / p
instead of the more accurate T = -ln(2) / ln(1-p) ?
The answer, I believe, is that the expression comes from:
dN/dt = - p N = - lambda N
which assumes continuous changes in N, not discrete changes.
Robert Cohen