Why can't a base be negative in an exponential function?

Question

The function $f(x) = a^x$ is generally taught to only allow $a > 0$. This is usually justified by giving a few examples of complex points in cases where $a < 0$. For example:
$f(x) = (-2)^x$, if $x = 0.5, f(x)=sqrt(-2)$, which of course is $0+sqrt(2)i$, a complex number. That being said, can't any number be approximated using odd denominators, and thus be put in the function? Example:
$x=3/4$ can be approximated using increasingly large powers of 3 in the denominator
$2/3<x<3/3$
$6/9<x<7/9$
$20/27<x<21/27$
and so on, in this way a real value of $f(x)$ can be always determined for any real $x$, even if $a<0$.
So why do teachers and graphing software refuse to consider $f(x)=(-e)^x$ (or similar) as functions?

Answer 1

Even though another answer has already been accepted, I'm posting this answer as I believe the previously accepted answer to be at least partially incorrect which is illustrated by the following table of evaluations

$$\begin{array}{cc} x & (-2)^x \\ 1 & -2. \\ 2 & 4. \\ \frac{3}{2} & 0.\, -2.82843 i \\ \frac{5}{3} & 1.5874\, -2.74946 i \\ \frac{8}{5} & 0.936764\, -2.88306 i \\ \frac{13}{8} & 1.18036\, -2.84963 i \\ \end{array}$$

whereas $(-2)^{\phi}\approx 1.11233\, -2.86093 i$.

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Also, the previously accepted answer didn't seem to specifically address the question which as I understand it is for example whether $\Re\left((-2)^{3/4}\right)$ could be approximated by $-(2)^{a/b}$ where $\frac{a}{b}$ is a rationale approximating $\frac{3}{4}$ and $b$ is odd. I believe this approach is misguided which is illustrated by the following table of evaluations

$\begin{array}{cccc} n & x=\frac{a}{3^n} & \frac{3}{4}-x & -2^x \\ 1 & \frac{2}{3} & \frac{1}{12} & -1.5874 \\ 2 & \frac{7}{9} & -\frac{1}{36} & -1.71449 \\ 3 & \frac{20}{27} & \frac{1}{108} & -1.67103 \\ 4 & \frac{61}{81} & -\frac{1}{324} & -1.68539 \\ 5 & \frac{182}{243} & \frac{1}{972} & -1.68059 \\ 6 & \frac{547}{729} & -\frac{1}{2916} & -1.68219 \\ 7 & \frac{1640}{2187} & \frac{1}{8748} & -1.68166 \\ 8 & \frac{4921}{6561} & -\frac{1}{26244} & -1.68184 \\ 9 & \frac{14762}{19683} & \frac{1}{78732} & -1.68178 \\ 10 & \frac{44287}{59049} & -\frac{1}{236196} & -1.6818 \\ 100 & \frac{386533140549008498277345847324215954526580641501}{515377520732011331036461129765621272702107522001} & -\frac{1}{2061510082928045324145844519062485090808430088004} & -1.68179 \\ \end{array}$

whereas $(-2)^{3/4}\approx -1.18921+1.18921 i$.

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For $a>0$, the function $f_a(x)=(-a)^x$ can be evaluated as

$$f_a(x)=a^x\, e^{i \pi x}=a^x\, (\cos(\pi x)+i\, \sin(\pi x))\,,\quad a>0\tag{1}$$

which leads to

$$\Re(f_a(x))=a^x\, \cos(\pi x)\,,\quad a>0\land x\in\mathbb{R}\tag{2}.$$

Answer 2

Let's consider the expression $(-2)^{\phi}$, where $\phi = \frac{1 + \sqrt2}{2} \approx 1.618034$ is the Golden Ratio. I picked $\phi$ because it has a conveniently-described series of fractional approximations: The ratio of consecutive Fibonacci numbers.

Let $\frac{p}{q}$ be a rational approximation of $\phi$. Then $(-2)^{\phi}$ can be approximated by $(-2)^{p/q} = \sqrt[q]{2^p}$. This gives us the approximation sequence

$$(-2)^{1/1} = -2$$ $$(-2)^{2/1} = 4$$ $$(-2)^{3/2} = \sqrt[]{-8} = -2.828427i$$ $$(-2)^{5/3} = \sqrt[3]{-32} \approx -3.174802$$ $$(-2)^{8/5} = \sqrt[5]{256} \approx 3.031433$$ $$(-2)^{13/8} = \sqrt[8]{-8192} \approx 2.849634 + 1.180357i$$

If you continue this series, you'll find that the magnitude of the numbers converges to $2^\phi \approx 3.069565$, but the direction ($\theta$) on the complex plane jumps around wildly. If $p$ happens to be even, you get a positive number for $2^p$ and thus a real number for $2^{p/q}$. But if $p$ is odd, then $2^{p/q}$ might be negative or complex. Which makes it hard to pin down the limit of the sequence.

Now, there are some calculators and programming languages that will let you just evaluate $(-2)^\phi$ directly.

>>> (-2)**1.618033988749895
(1.1123331010556903-2.8609336481188627j)

But understanding that requires knowing calculus and Euler's formula and branch cuts, as I've explained in my most popular answer on this site. And as far as I can tell from your comments, you're only in Algebra II or Precal. So your teacher is just giving the standard response to problems that require math you haven't learned yet: Don't do that.