log1p(x) returns natural logarithm of 1+x (base e )( What is e ? )
import math
print(math.log1p(2))  # 1.0986122886681098
print(math.log1p(4))  # 1.6094379124341003
print(math.log1p(0))  # 0.0 
Using negative number
For any value less than or equal to 0 , we will get ValueError
import math
Above code will generate error.

log modf()


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