This proposition is a lemma related to another stage for defining exponential function $a^{x}$, in this case for reals, taking into account it is defined for rationals.
Proposition Let $a>1$ and $A, A^{*}$ be sets given by \begin{align*} A &=\{a^{r}\mid r<x;\, r\in\mathbb{Q}\}\\ A^{*} &=\{a^{s}\mid x<s;\, r\in\mathbb{Q}\} \end{align*} Then $\sup A=\inf A^{*}$.
Proof (Partial). We need to show that this sets are not empty and they are properly bounded. Let $\epsilon>0$, then $x-\epsilon<x<x+\epsilon$. By the density of the rationals in reals, there are $r',s'\in\mathbb{Q}$ such that \begin{align*} x-\epsilon<r'<x<s'<x+\epsilon \end{align*} therefore \begin{align*} a^{r'}\in A, &a^{s'}\in A^{*}\therefore\\ A\neq\emptyset, &A^{*}\neq\emptyset \end{align*}
Using the same $r',s'$ from the former argument we can see that \begin{align*} a^{r'}\notin A^{*},\,a^{s'}\notin A \end{align*} since \begin{align*} r'<x<s' \end{align*} By the last inequality \begin{align*} a^{r'}<a^{s'} \end{align*} since $a>1$. But \begin{align*} a^{r}<a^{s'},\,\forall r<x<s' \end{align*} therefore $a^{s'}$ is an upper bound for $A$. An analogous argument shows that \begin{align*} a^{r'}<a^{s},\,\forall r'<x<s \end{align*} therefore $a^{r'}$ is a lower bound for $A^{*}$. By the supremum principle $A$ has supremum and $A^{*}$ has infimum.
We have $a^{r}<a^{s}$ for every $r<x$ and $x<s$. Then every $a^{s}$ is an upper bound for $A$. That is $\sup A\leq a^{s}$ by definition of supremum. But this means $\sup A$ is a lower bound for $A^{*}$. We conclude that $\sup A\leq\inf A^{*}$.
But, how to discard the option $<$? I can't see it! What is sure, is that is impossible to be proved seeking for the reversed inequality, for always we have $a^{r}<a^{s}$. The proof can continue trying to prove that $\sup A+\epsilon$, with $\epsilon>0$ is not a lower bound for $A^{*}$, what is the same as proving $\inf A^{*}-\epsilon$ is not an upper bound for $A$. The problem is how to write this in such a manner to compare with things like $a^{r}$. Could anyone can help?
Suppose $\sup A$ is strictly less than $\inf A^*$, then there is a $y \in \mathbb{Q}$ such that $\sup A < y < \inf A^*$. But either $x<y$ causing $y \in A^*$, $x = y$ causing $x \in \mathbb{Q}$, or $x>y$ causing $y \in A$. By supposition, $y > \sup A$ and $y < \inf A^*$, so neither $x>y$ nor $x<y$. Therefore $x=y$ and $x \in \mathbb{Q}$. This means the only rational number missing from $A \cup A^*$ is $x$ and we must have $\sup A = x = \inf A^*$.