As you noted, the rational exponents can be accounted for. But, the rationals densely pack the number line, and (more importantly) the exponential function, restricted to rationals, is continuous. For instance, the limiting value of $2^x$ as $x → 0$ through rational numbers is $2^0 = 1$:
$$2^{1/2} = 1.4142135623⋯, \hspace 1em 2^{1/4} = 1.1892071150⋯, \hspace 1em 2^{1/8} = 1.0905077326⋯, \\
\hspace 1em 2^{1/16} = 1.0442737824⋯, \hspace 1em 2^{1/32} = 1.0218971486⋯, \hspace 1em 2^{1/64} = 1.0108892860⋯, \\
\hspace 1em 2^{1/128} = 1.0054299011⋯, \hspace 1em 2^{1/256} = 1.0027112750⋯, \hspace 1em 2^{1/512} = 1.0013547198⋯, \\
2^{1/1024} = 1.0006771306⋯, \hspace 1em 2^{1/2048} = 1.0003385080⋯, \hspace 1em ⋯.
$$
If a function is continuous over the rationals, then because the rationals densely pack the number line, there is a unique extension of the function to a continuous function defined over the whole number line.
So, at some point, that extra assumption is stipulated as an axiom: that the exponential function $x ↦ 2^x$ be continuous for all real numbers; and the exponential function is defined as the unique continuous extension of the function from rational numbers to all real numbers.
You can go a step further: you can also assume that the graph $y = 2^x$ has a well-defined slope as it passes through $(x,y) = (0,1)$. Call this slope $ln2$. For the example just tabulated:
$$
2^{1/128} = 1 + \frac{0.6950273⋯}{128}, \hspace 1em
2^{1/256} = 1 + \frac{0.6940864⋯}{256}, \hspace 1em
2^{1/512} = 1 + \frac{0.6936165⋯}{512}, \\
2^{1/1024} = 1 + \frac{0.693381⋯}{1024}, \hspace 1em
2^{1/2048} = 1 + \frac{0.693264⋯}{2048}, \hspace 1em,
⋯
$$
Thus $ln2 = 0.693⋯$. So, there is a simple process to tabulate $2^x$:
- Tabulate $y = 2^x$:
- ⓿: Start with $x$.
- ❶: Halve: $x ← ½x$, $n$ times, for some large number $n$.
- ❷: Set $y = 1 + x×ln2$.
- ❸: Square: $y ← y^2$, $n$ times.
- ❹: End up with $y ≃ 2^x$
The number of times,
$n$, is chosen large enough to get accurate results (the accuracy ratio being approximately given by the result at the end of ❶), but not so large that the rounding errors swamp everything out.
To find $ln2$, go in the opposite direction, but with a slope of $1$, instead of $ln2$:
- Tabulate $x = ln2$:
- ➃: Start with $y = 2$.
- ➂: Square root: $y ← \sqrt{y}$, $n$ times, for some large number $n$.
- ➁: Set $x = y - 1$.
- ➀: Double: $x ← 2x$, $n$ times.
- ⓪: End up with $x ≃ ln2$.
There is also a unique number $e$ the slope of whose graph $y = e^x$ is 1 as it crosses the point $(x,y) = (0,1)$. It's found by repeating the same process used to find $y = 2^x$, but using the slope $lne = 1$ in place of $ln2$.
Since the first process is equivalent to starting with $x×ln2$, instead of with $x$ and replacing step ❷ with a slope of 1: $y = 1 + x$, then it follows that $2^x = e^{x×ln2}$ and, therefore, that $2 = e^{ln2}$.
So, the graphs of all the exponential functions $y = b^x$, for $b > 0$, are just copies of the graph $y = e^x$, rescaled as $y = e^{x×lnb}$, where $lnb$ is the unique exponent of $e$ for which $b = e^{lnb}$.
To find $e$, itself, repeat the process used to tablulate $y = 2^x$, but starting with $x = 1$ and using its slope $lne = 1$ in place of $ln2$. The result will be $y = e^{x×lne} = e^{1×1} = e^1 = e$.
But, wait ... there's more!
Since
$$\cos{2x} = \cos^2{x} - \sin^2{x}, \hspace 1em \sin{2x} = 2 \sin{x} \cos{x},$$
then it follows that
$$\left(\cos{x} + i \sin{x}\right)^2 = \cos{2x} + i \sin{2x},$$
where $i^2 = -1$. The graph for $y = \cos{360x°} + i\sin{360x°}$ also passes through $(x,y) = (0,1)$. It is also continuous at $(x,y) = (0,1)$:
$$\begin{align}
\cos 360° + i \sin 360° &= ½ \left(2 + i×0\right) \\&= 1, \\
\cos 180° + i \sin 180° &= ½ \left(-2 + i×0\right) \\&= -1, \\
\cos 90° + i \sin 90° &= ½ \left(0 + i×2\right) \\&= i, \\
\cos 45° + i \sin 45° &= ½ \left(\sqrt{2} + i×\sqrt{2}\right) \\&= 0.7071067⋯ + i×0.7071067⋯, \\
\cos 22½° + i \sin 22½° &= ½ \left(\sqrt{2 + \sqrt{2}} + i×\sqrt{2 - \sqrt{2}}\right) \\&= 0.9238795⋯ + i×0.382683⋯, \\
\cos 11¼° + i \sin 11¼° &= ½ \left(\sqrt{2 + \sqrt{2 + \sqrt{2}}} + i×\sqrt{2 - \sqrt{2 + \sqrt{2}}}\right) \\&= 0.9807852⋯ + i×0.195090⋯, \\
\cos 5⅝° + i \sin 5⅝° &= ½ \left(\sqrt{2 + \sqrt{2 + \sqrt{2 + \sqrt{2}}}} + i×\sqrt{2 - \sqrt{2 + \sqrt{2 + \sqrt{2}}}}\right) \\&= 0.9951847⋯ + i×0.098017⋯, \\
⋯
\end{align}$$
So, the function $y = \cos{360x°} + i\sin{360x°}$ also behaves like an exponential which - since $y = 1$ when $x = 1$ - we will simply call $y = 1^x$, by fiat! We'll make it an explicit exception to $y = b^x$ process, since that would have given us $y = 1$, if $b = 1$. Unlike the exponentials $b^x$, $1^x$ is periodic: $1^{x+1} = 1^x$; and it is, in fact, equal to $1$ when $x$ is an integer.
So, what's its slope? That is: what is $ln1$?
$$
\cos 360/16° + i \sin 360/16° = 1 + (-1.217927⋯ + i×6.122934⋯)/16, \\
\cos 360/32° + i \sin 360/32° = 1 + (-0.614871⋯ + i×6.242890⋯)/32, \\
\cos 360/64° + i \sin 360/64° = 1 + (-0.308177⋯ + i×6.273096⋯)/64, \\
⋯
$$
It approaches a multiple of $i$ in the limit that we will call $2π ≅ 6.2⋯$. Therefore, $ln1 = 2πi$.
Therefore, the same process applies as before. But since not all calculators come natively-equipped with complex arithmetic (even though computers now do, since programming languages, like C and C++, have made it part of their respective standards), then we'll adapt it to the following form:
- Tabulate $c + is = 1^x$, where $-½ < x < +½$:
- ⓿: Start with $x$.
- ❶: Halve: $x ← ½x$, $n$ times, for some large number $n$.
- ❷: Set $(c,s) = (1,2πx)$. Since $c^2 + s^2$ is supposed to be 1, then adjust it by dividing $c$ and $s$ by $\sqrt{c^2+s^2}$.
- ❸: Square: $(c,s) ← (c^2 - s^2,2sc)$, $n$ times.
- ❹: End up with $c + is ≃ 1^x$
Since it's periodic in
$x$, then values outside the range
$-½ < x < +½$ can be adjusted by subtracting the nearest integer to
$x$ from
$x$. For the half-integers,
$x = ⋯, -2½, -1½, -½, +½, +1½, +2½, ⋯$, we already have
$1^x = -1$, so nothing further needs to be done in those cases.
Likewise, as we did with $ln2$, to find $ln1$, we adapt the second procedure:
- Tabulate $x = ln1$:
- ➃: Start with $y = -2$.
- ➂: Square root: $y ← \sqrt{2 + y}$, $n-1$ times, for some large number $n$.
- ➁: Set $(c,s) ← (0,\sqrt{2 - y})$.
- ➀: Double: $s ← 2s$, $n$ times.
- ⓪: End up with $c + is ≃ 2πi$; i.e., with $c = 0$ and $s = 2π$.
