How simple is it to prove the statement that for each $a,b,x \in \mathbb R$, we have $x^a \cdot x^b = x^{a+b}$, using Dedekind cuts? The reason I'm curious about this question is that someone was trying to solve $(\sqrt 2)^2 = 2$ using Dedekind cuts, and I immediately saw this is implied directly by the statement above.
So, using Dedekind cuts, how easy is it to prove the first statement? The first statement is obviously true "in the long run", but my worry is that the proof will require machinery to the level of forgetting that we are even working with Dedekind cuts in the first place and not just over $\mathbb R$.
To solve this statement typically, I think I would first define the map $x \mapsto e^x$ using limits, and show $e^{x+y}=e^x e^y$ by matching up the series expansions for each degree of the power series on each side. Then the rest should follow by $e^{\log b \cdot c} = b^c$ and so $x^a x^b = e^{a \log x}e^{b \log x}=e^{(a+b) \log x} = x^{a+b}$.
However, how can this proof be adapted to please someone who deals with the real numbers only through the interface of Dedekind cuts?
