using math.log(var) in python

Question

I think if I'd know the math behind it I might have figured it out by myself.

When I call math.log10(0.0000000000001) I get -13. But how do I convert this back into 0.0000000000001? and how i'd handle it when using math.log(0.0000000000001)?

Both ways, the fundamental, arithmetic one and the one using a built-in python function would interest me.

score 2 · Answer 1 · answered Dec 02 '15 at 22:43

2

a = math.log10(0.0000000000001)
b = 10 ** a

answered Dec 02 '15 at 22:43

Daniel

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sorry, i edited my question. what is when math.log() is used instead of math.log10 ? – bngschmnd Dec 02 '15 at 22:51
1

@bngschmnd - math.log uses base e, instead of base 10. So, math.exp() is the inverse of math.log() – user1245262 Dec 02 '15 at 22:55

score 0 · Answer 2 · answered Dec 02 '15 at 23:21

math.log() takes two parameter, a number and a base. Default value is e (mathematical constant), a special constant which tends to 2.71828.

You can get the number by evaluating base to the power log_value

>>> a = math.log(10)
>>> a
2.302585092994046
>>> 2.71828**a
9.999984511610785

If I would take e's value with higher precision, I can get the result close to 10.

Or, you can use math.exp()

>>>math.exp(a)
10.000000000000002

You can also use math.pow()

>>> math.pow(2.71828,a)
9.999984511610785

using math.log(var) in python

2 Answers2