Fix $b>1$. Let $B(x) = \{b^t : t \in \mathbb{Q}\text{ & }t≤x\}$ and let $B'(x) = \{b^t : t \in \mathbb{Q}\text{ & }t<x\}$.
Show that $\sup B(x) = \sup B'(x)$. It is quite easy to show the result holds for any irrational $x$. Yet I've been stuck for hours trying to prove it for rational $x$. It's also not difficult to show that $\sup B(x) ≥ \sup B'(x)$.
From here an idea would be to assume $ \alpha = \sup B(x) > \sup B'(x)$ and reach a contradiction. Which would be easy by finding $r \in \mathbb{Q}$ such that $ \alpha < b^r < b^x$. The only problem is that I'm not allowed to use any material that isn't in the first chapter of Baby Rudin.
I would really appreciate any help.
If $\ x\ $ is rational, then $\ \sup B(x) = b^x\ $. If $\ \epsilon\ $ is any positive real number with $\ \epsilon < b^x\ $, let $\ d\ = \frac{1}{\left(1-\frac{\epsilon}{b^x}\right)}-1 $, $\ q\ $ be an integer greater than $\ \frac{b}{d} $, and $\ t= x-\frac{1}{q}\ $. Then $\ \left(1 + d\right)^q \ge 1 +q\,d > b\ $, so $\ \frac{1}{\left(1-\frac{\epsilon}{b^x}\right)}=1+d > b^\frac{1}{q}\ $. Inverting this gives $b^{-1/q} > 1-\frac{\epsilon}{b^x}\ $. Thus we have $\ b^t \in B'\left(x\right)\ $, $\ b^t = b^{x-\frac{1}{q}} > b^x - \epsilon\ $, and it follows that $\ \sup B'(x) > b^x - \epsilon\ $. Since $\ \epsilon\ $ can be chosen arbitrarily close to $\ 0\ $, it follows that $\ \sup B'(x) \ge b^x $.
Amendments and notes
I have edited this proof to eliminate the use of logarithms, which a comment below informs us have not been covered at the point when the requested proof is asked for. There appear to me to be only two steps in the above proof that might be considered problematic: