Why are logarithms a function while exponentials are operations This might sound a bit like a stupid question and probably is, but I have been wondering for some time now why is it that if logs are meant to be the inverse of exponentiation why are they treated as functions while exponentials are treated as an operation?
Is there some historic reason for this?
 A: This is a good question. Historically, exponentiation is a binary
operation similar to addition and multiplication except it is not 
commutative or associative, and the base is a positive real while
the exponent is any real number.
The logarithm is also a kind of binary operation except the base of logarithms is a positive real number usually fixed by convention to 
$10$ or $e$. When this is done, the logarithm operation becomes a 
function of one variable. To emphasize this, there are extensive
tables of logarithms to base $10$ usually known as "log tables". 
There are only rare corresponding tables of the exponential.
What is usually done in practice is to use a log table to find the 
inverse exponential. This is known a finding the "antilog" of a
 number using a log table. For details read the
Wikipedia article Mathematical table. In summary, it is just a historic
 accident in the way logarithms and exponentials are computed
 using function tables.
Extending now to complex numbers, an important fact is that the exponential function to base $e$ which is denoted by $\exp(z)$ is
an entire analytic function of the complex variable $z$. However,
when we write $e^z$ as an exponentiation operation, the conventional
definition makes this multi-valued just like any other
$b^z := \exp(\log(b)z)$ because the logarithm is multi-valued for 
complex numbers. Thus, by the conventional definition, $e^z$ as a
binary operation of exponentiation is not the same as $\exp(z)$
regarded as a one variables function. For the
details you can read the DLMF sections 4.2(iii) and 4.2(iv) on exponentials
and powers. Of course, if you use the principal branch
of the logarithm $\,\log(b)\,$, then $\,b^z\,$ becomes
an entire analytic function of $\,z\,$ also.
Notice, however, that $\,z^c\,$ where $\,c\,$ is some
fixed complex number is, in general, a multi-valued function of $\,z\,$. This is different than the previous case of $\,b^z\,$
where the base was fixed. Here, the power is fixed and the base
is variable. If the exponent $\,c\,$ is an integer,  then the
function is a rational function of $\,z\,$ and, aside from a pole
at $\,0\,$ if $\,c<0\,$ it is an analytic function. Thus the binary
operation of exponentiation leads to two different kinds of functions
by fixing one of the operands. One of the notable consequence of this, is
that, in general, the value of $\,b^z\,$ is $\,1\,$ if $\,z=0,\,$
while the value of $\,z^c\,$ is $\,0\,$ if $\,z=0\,$ assuming that
$\,c>0.\,$ This leads to the well-known question of what is the
value of $\,0^0?\,$ The answer depends on convention and context.
