Is it possible to ensure that the following characterisation of the function $a^x$ is well-defined? Say we define $a^x$ in the obvious way for $x\in\mathbb{Q}$, with $a^{p/q}=\left(\sqrt[q]{a}\right)^p$. This definition can be extended to irrational numbers by requiring that $a^m<a^x<a^n$, where $m$ and $n$ are rational numbers such that $m<x<a$.* How can we prove that this definition is enough to narrow down the value of $a^x$ to just one number? Moreover, does this definition automatically imply that the function $x \mapsto a^x$ is continuous?
Ideally, I would look to see a solution that doesn't rely on other characterisations of the exponential function, e.g. when $a^x$ is defined as $\exp(x\log(a))$.

*This definition only applies when $a>1$. If $a<1$, then the inequalities are reversed.
 A: It's best to define $x\mapsto a^x$ on irrational $x$ by demanding continuity, rather than making inequalities' directions depend on whether $a>1$, with $1^x:=1$. In other words, $a^x:=\lim_{n\to\infty}a^{x_n}$ for any sequence $x_n$ in $\Bbb Q$ with $x=\lim_{n\to\infty}x_n$. Then you just need to check the choice of $x_n$ is irrelevant; this is equivalent to proving all null sequences (those with $x=0$) give $\lim_{n\to\infty}a^{x_n}=1$. Of course, once this is done, your inequality-based definition is trivially equivalent.
Edit, for requested elaboration:
Suppose $y_n,\,z_n$ are rational sequences with common limit $L$. Let $\delta_n:=y_n-z_n$, so $\lim_{n\to\infty}\delta_n=0$; we say $\{\delta_n\}_{n\ge0}$ is a null sequence. If we can prove this is sufficient to force $\lim_{n\to\infty}a^{\delta_n}=1$,$$\frac{\lim_{n\to\infty}a^{y_n}}{\lim_{n\to\infty}a^{z_n}}=\lim_{n\to\infty}a^{\delta_n}=1\implies\lim_{n\to\infty}a^{y_n}=\lim_{n\to\infty}a^{z_n},$$so $a^L$ is well-defined. (Well, we also need to check $\lim_{n\to\infty}a^{y_n},\,\lim_{n\to\infty}a^{z_n}$ exis, so the division is legal.) However, that null sequences satisfy $\lim_{n\to\infty}a^{\delta_n}=1$ is just the special case of the desired result where the exponents tend to $0$, so the general case reduces to that special case.
A: (Assuming $a>1$), you can perhaps let
$$ a^x=\inf\{\,y\in\Bbb R\mid \forall p\in\Bbb Z\colon\forall q\in\Bbb N\colon qx>p\to y^q>a^p\,\}.$$
