Consider some $a \in \mathbb{R}$ and $x \in \mathbb{R}\backslash \mathbb{N}$.
Is there some intuition to be had for the number $a^x$?
For example the intuition of $a^2$ is obvious; it's $a*a$ which I can think about with real world objects such as apples (when $a \in \mathbb{N}$). What about $a^{1.9}$?
Having defined positive integer exponents, if you want the property $$a^m \cdot a^n= a^{m+n}$$ to continue to hold, then you must define $a^0=1$ and $a^{-n} = 1/a^n$ for integer $n$. This takes care of all integers. Then, if you want the property $$a^{mn} = (a^m)^n$$ to continue to hold, you must define $a^{p/q} = \sqrt[q]{a^p}$ (for positive real $a$, and integers $p$ and $q$, $q \ne 0$). This takes care of all rational numbers. And then, if you want the function $a^x$ to be continuous from $\mathbb{R} \to \mathbb{R}$, there is only one such extension.
If we fix $a>0$, $f(x)=a^x$ is continuous on $\mathbb R$. The intuition behind rational exponents is pretty clear, and one extends from the rationals to all reals in this manner.
Yet another intuition follows from the following formula: $$x^{\,y} = e^{\;y\log x}$$ (with appropriate restrictions of course).
It has to do with extending exponentiation to rational number exponents using various properties of multiplying powers and powering powers.
We know that for positive real $r$,
$$(r^a)^b=r^{ab}$$
Without loss of generality, let $z$ be an integer.
Using the law of powers of powers.
$$(r^{1/z})^z=r^{1z/z}=r^1=r$$
Taking the ($z$)th root of the equation above.
$$r^{1/z}=\sqrt[z]r$$
Let $z_1$ and $z_2$ be integers.
$$(r^{1/z_1})^{z_2}=r^{z_2/z_1}$$
As such, we can well-define arbitrary rational exponents for positive reals. A positive real raised to an arbitrary rational power has exactly one positive real value for each rational number.
And because all real numbers are arbitrarily approximated by rational numbers and can be ordered among the rational numbers, arbitrary real powers can likewise be approximated. We can define $\pi^{\pi}$ and even conclude it is between $\pi^{3.141}$ and $\pi^{3.142}$.
Going beyond positive real bases involves $i^2=-1$, natural logarithms, natural antilogs, and trigonometry.
