# Assumptions in mathematica when Symplify on Log[Exp[a/b]]

In mathematica 9.0:

This:

Assuming[b > 0 || b < 0, Simplify[Log[Exp[a/b]]]]


yields:

a/b


However, this:

Assuming[FullSimplify[b > 0 || b < 0], Simplify[Log[Exp[a/b]]]]
enter code here


or this:

Assuming[b !=0 , Simplify[Log[Exp[a/b]]]]


Yield:

Log[Exp[a/b]]


Shouldn't be both equivalent? Why isn't the expression simplified in the second case?. Also tried to add ass an assumption a and b to be real, just in case, but doesn't work either. Could someone shed any light on this?

• I evaluated Assuming[b > 0 || b < 0, Simplify[Log[Exp[a/b]]]] in Mathematica ver.7, 8 and 9 runing Windows. I get always Log[Exp[a/b]]], never a/b as you claim. What version do you use? – Artes Oct 16 '13 at 19:55
• FullSimplify[Log[Exp[a/b]], Assumptions -> {a \[Element] Reals, b > 0}] gives a/b in v9 but original formulation gives Log[Exp[a/b] – bill s Oct 16 '13 at 20:09

I suspect that this has to do with the assumption b > 0 implicitly stating that b is a real, where that is not the case with b != 0. This is, of course, because there is no natural ordering of complex numbers, but they do have a 0-element to compare against. As evidence for this answer, note that
Assuming[Element[b, Reals] && b != 0, Simplify[Log[Exp[a/b]]]]

Gives the expected result of a/b where it didn't above.
• This is exactly right. Try FindInstance[Log[Exp[a/(1 + I)]] != a/(1 + I), a] – Chip Hurst Oct 16 '13 at 21:09