Sort of.
First, we have to ask what $f(x) = a^x$ means for $f: \mathbb{R} \to \mathbb{R}$, where $a \in \mathbb{R}$ is a constant. Assuming that $a > 0$, we can write:
$$
f(x) = a^x = \left( e^{\ln(a)}\right) ^x = e^{(\ln a)x},
$$
so it boils down to having a sensible definition of the natural exponential and logarithm functions.
The exponential function can be defined by its Maclaurin series:
$$
e^x = \sum_{n=0}^\infty \frac{x^n}{n!} = 1 + x + \tfrac{1}{2}x^2 + \tfrac{1}{6}x^3 + \cdots
$$
One has to, of course, verify that this infinite series converges for every $x \in \mathbb{R}$ for this definition to make sense.
The same series can be used to define the matrix exponential function, interpreting $x$ now as an $n \times n$ matrix. (It also converges for all matrices.)
But, here the analogy begins to fall apart.
First of all, the matrix logarithm is a much more finicky creature. The standard Mercator series for the logarithm doesn't converge for all matrices. Even if we extend our ground field to the complex numbers, then a matrix has a logarithm if and only if it's invertible. Even when a logarithm exists, it is not unique!