Number system with $e^x = 0$ for some $x$ It is well known that $e^x \ne 0$ for all $x \in \mathbb{R}$ as well as $x \in \mathbb{C}$.  Upon reading this article and doing a bit of research I have found that this also applies to the quaternions $\mathbb{H}$, the octonions $\mathbb{O}$ as well as the space of $m$ by $n$ matrices with real or complex entries.
My question is whether there is ANY number system at all for which $e^x = 0$ for some $x$, that is, $\log 0$ is defined and has a finite value.  Preferably the example should be finite-dimensional and should not be constructed by arbitrarily assigning a value to $\log 0$, such as $\log 0 := 42.$
Additionally, for the purposes of this question, none of the usual properties of arithmetic or the exponential function are assumed true, though I suppose this makes my question somewhat meaningless.  
Edit: I am intrigued at Yuriy S's idea of defining $e^x = 0$ for all $x$. My question now is what is the most "well behaved" algebra we can come up with if $e^x$ is required to be identically zero? 
 A: The floating-point number system (for a given number of bits) is a finite subset of the extended real number line. It relaxes various algebraic identities so that it can remain closed under as many operations and inputs as possible. As you put it, the "usual properties of arithmetic or the exponential function" are not true in this system, although they are approximately true for the most part.
In this system, $-\infty$ is a number and $e^{-\infty}=0$.
References:
http://pubs.opengroup.org/onlinepubs/9699919799/functions/exp.html
http://en.cppreference.com/w/c/numeric/math/exp
A: If you want the properties
$$e^{x+y}=e^xe^y\quad\hbox{for all $x,y$}$$
and
$$e^0=1$$
to remain true, then we have
$$e^xe^{-x}=1\quad\hbox{for all $x$}$$
and so $e^x$ can never be zero.
A: Adding to what David said, even if you give up on the homomorphism part of the exponential, the series definition (which holds in any Banach algebra, which includes the number systems) make so that $e^X e^{-X}=1$, so $e^X$ can't be $0$.
