(12C) log1p function

[Albert Chan]

This trick is from John D Cook's Tricks for computing log(1+X)

I picked the formula from the comment section instead.
The result is slightly more accurate, and no worry about divide by zero.

log1p(X) = ln(1 + X) - (X + 1 - 1 - X) / (X + 1)

Example: log1p(X = 1.23456789e-6)

1.23456789 [EEX] 6 [CHS] [Enter] ; X value

[Enter] 1 + 1 - [X<>Y] - [CHS]     ; epsilon => -4.3211e-10
[Lst-X] 1 + ÷        ; correction to epsilon => -4.321094e-10
[Lst-X] [LN]          ; LN(1+X) => 1.234999237e-6
+                         ; log1p(X) => 1.234567128e-6 (all digits correct)