Ttdfjt

Ok, this is awkward.

E^Log[a]

gives a, which is what I expected, BUT,

Log[E^a]

is not evaluated at all (I expected the result to be a—this is actually an MWE, what I was trying to do was more complex).

Since I still struggle to understand some of Mathematica's concepts (like when expressions are/n't evaluated), I thought (wrongly, it seems) that it was an Evaluate or Assumptions issue, but now I think it wasn't. I tried:

Assuming[a ∈ Reals, Log[E^a]] // Evaluate
Assuming[a > 0, Log[Exp[a]]] // Simplify

and other variants—which in turn shows that I'm giving a shot in the dark.

What is the explanation for this behavior? How do I get to a? Am I missing an assumption? I don't really care about any weird special cases, a is real, not complex, nor a function, nor anything else.

It obviously seems that I'm missing something (possibly very simple and obvious), but I can't figure out what.

Since Log is a multi-valued inverse function of Exp, Mathematica doesn't evaluate Log[Exp[a]] (or equivalently, Log[E^a]). If you want to simplify these, you need to provide an assumption on the domain of a, e.g.,

Simplify[Log[E^a], a ∈ Reals]

a

or

Simplify[Log[E^a], a > 0]

a

as suggested in the other answer. Another method is to use PowerExpand:

PowerExpand[Log[E^a], Assumptions -> True]

a + 2 I π Floor[1/2 - Im[a]/(2 π)]

Addendum

As an aside, Log behave exactly like ArcSin here:

ArcSin[Sin[x]]

ArcSin[Sin[x]]

Including a domain restriction:

Simplify[ArcSin[Sin[x]], -Pi/2 < x < Pi/2]

x

For larger domains, Simplify doesn't work:

Simplify[ArcSin[Sin[x]], 0 < x < 2 Pi]

ArcSin[Sin[x]]

Again, using PowerExpand is useful:

p = PowerExpand[ArcSin[Sin[x]], Assumptions -> 0 < x < 2Pi];
p //TeXForm

$begincases
pi -x & fracpi 2<xleq frac3 pi 2 \
x & xleq fracpi 2 \
x-2 pi & textTrue
endcases$

Visualization:

GraphicsColumn[
 Plot[ArcSin[Sin[x]], x, 0, 2 Pi],
 Plot[p, x, 0, 2 Pi]
]

The assumption a > 0 is needed when Simplify is called:

Assuming[a > 0, Log[Exp[a]] // Simplify]

a

Logarithm is a problematic function for MMA. I do not know the reason. One can see, for example Log[a] + Log[b] // Simplify returns Log[a] + Log[b] though LeafCount[Log[a] + Log[b]] yields 5, while LeafCount[Log[a*b]] gives 4.

I do not think that in the case Log[E^a] we should use such conditions as a>0 or whatever, since the relation in question is universal. On the other hand, in the case Log[E^a]===a the expression a is doubtless much simpler and, therefore, preferable. It is not that clear what one prefers to have in the case Log[a] + Log[b]

The way around can be using rules. For example, in this case one may use

rule = Log[x_^y_] :> y*Log[x];

Then its application gives

In[5]:= Log[E^a] /. rule

 (* Out[5]= a *)

Have fun!