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While tutoring a student recently, I have come across the situation of explain logarithms by first introducing functions of the form $$f(x)= a^x$$ where $a \ge 0,x\in \mathbb{R}$. My student then asked me a seemingly innocent question, what if $a < 0.$ I explained to her, that it suffices to consider the basic case $$g(x)= (-1)^x$$ however I told her that the answer to that is beyond the scope of the material. I did tell her the answer though, namely that $$g(x) = (-1)^x = (e^{i \pi})^x = e^{i x \pi} $$ which is the unit circle in the complex plane. Of course this opens up a new can of worms.

This got me to thinking, in the course of history the most probable chronological sequence was that functions of the form of $f(x)$ were considered before complex numbers were formalized. Then did mathematicians of the past simply state that functions with $a<0$ were simply undefined? Or was it the case that complex numbers were considered first?

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This is not an answer to your question, but I think you blew it in your answer to the student: There's a countably infinite number of reasonable answers to her question! $(-1)^x=e^{(2k+1)\pi x}$ for any integer $k$. – Harald Hanche-Olsen Nov 9 '09 at 22:24
Er, forgot the factor $i$ in the exponent. Too bad you can't preview comments. – Harald Hanche-Olsen Nov 9 '09 at 22:26
Ever hear of occam's razor Harald ? :P – user1447 Nov 9 '09 at 22:43
Occam's razor? Yes, but I fail to see its relevance here. – Harald Hanche-Olsen Nov 10 '09 at 2:09
"entities must not be multiplied beyond necessity", thats all. – user1447 Nov 11 '09 at 0:54

Rafaele Bombelli in l'Algebra (1572) considered imaginary roots of cubic equations as numbers since they might cancel out each other in the course of a following calculation. So he was the first who dared to use imaginary numbers in calculations. How much this calculations were "formalized" is a matter of opinon. Certainly this was not a formalization in modern sense. However, from Pythagoras over Archimedes to Gauß formalization has been sufficient to obtain correct results. Therefore Bombelli is at least one of the inventors of imaginary numbers. Sometimes also Cardano is nominated.

This knowledge spread soon. Albert Girard in Invention nouvelle en l'algèbre (1629) treated roots of negative numbers already without any ado. He distinguished the roots 3 and -3 of 9 and said of the root of -9 that it cannot be decided as positive or negative.

René Descartes baptized these new numbers in 1637. In his Géométrie he talks about imaginary roots (radices imaginariae) of an equation in contrast to true and false roots (radices vera, radices falsae) the latter denoting negative numbers, which was a common expression at that time.

As the inventors of logarithms we can name Jost Bürgi (his Arithmetische und geometrische Progresstabulen were published in 1620, but discovered already about 1590), and John Napier whose Descriptio appeared in 1614.

The word function, not yet used in modern sense (even Euler required a single formula to define a function), has been invented by Leibniz in his Methodus tangentium inversa, seu de fuctionibus (1673).

So the original question can be answered: First there were imaginary numbers, later came functions and logarithms. And logarithms of negative numbers came last.

Leibniz himself denied the existence of such logarithms and had a controversy with Johann Bernoulli about that topic in an exchange of letters in 1712 to 1713. Bernoulli defended his logarithme imaginaire. Finally Euler solved the problem in full generality. But that is a well known story.

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Short answer: Most likely undefined.

Long answer: The "naive" definition of $f(x) = a^x$ where $a, x \in \mathbb{R}$ and $a > 0$ is as follows. You know how to define $f(n)$ where $n$ is an integer. You can define $a^{p/q}$ to be the unique positive real number satisfying $(a^{p/q})^q = a^p$, so you know how to define $f(n)$ where $n$ is rational. When $a > 0$, it turns out that $f$ is continuous on $\mathbb{Q}$, so one can sensibly define $f(x)$ for all real $x$ by finding a Cauchy sequence of rationals converging to $x$ and taking the limit.

When $a < 0$, this reasoning falls apart at the second step: there is no positive real number satisfying $(a^{p/q})^q = a^p$ if $p$ is odd and $q$ is even. Even at the other rational numbers, the resulting function is highly discontinuous. So the only sensible thing to do, from a real-variable perspective, is to ignore this case.

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thanks for the answer, I suspect your reasoning is correct. Correct me if I am wrong but f continuous on isn't a sufficient condition. For example, take $f(x) = 1 \text{ for } x \in \mathbb{Q}$ and $f(x)=0 \text{ for } x \in \mathbb{R} \ setminus \mathbb{Q} $ then that method would imply f(x) is 1 everywhere, since f(x) is continuous on the rationals – user1447 Nov 9 '09 at 23:03
A continuous function is uniquely determined by its values on a dense subset, so continuous functions on Q extend uniquely to continuous functions on R. The function you wrote down isn't that unique extension. – Qiaochu Yuan Nov 9 '09 at 23:04

I have stumbled upon this article, maybe it can help:

History of the Exponential and Logarithmic Concepts, Florian Cajori, The American Mathematical Monthly, Vol. 20, No. 7 (Sep., 1913), pp. 205-210.

By the way, already $(-1)^{1/2}$ is undefined/multivalued.

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