Talk:Existence of Logarithm

The uniqueness proof could be a bit more detailed. As I understand it, it implicitly uses $b^x = 1 \implies x = 0$, or at the very least, that $b^x = 1$ has atmost one solution. Thus a small reference to the fact that $b^x > 1$ whenever $x > 0$ would be great.

Agreed -- this proof is already up for restructuring. --prime mover (talk) 13:26, 26 April 2015 (UTC)