Has a dyadic logarithm been studied? Define a dyadic logarithm of an odd dyadic integer $a$ as the limit of $\frac{a^z-1}{z}$ for $z\to0$ in the dyadic sense. I can prove, I'm assuming it's also well known, that dyadic exponentiation makes sense. (This is not true in general for p-adic integers.) A definition that works is to define $a^z$ as the limit of $a_n^{z_m}$ where $a_n$ is a sequence of positive integers that converges to $a$ in the dyadic sense and similarly $z_m \to z$.
I also have a proof that what I'm calling the dyadic logarithm exists and has some algebraic properties, such as the dyadic logarithm of a product is the sum of the dyadic logarithms. The dyadic logarithm of $-1$ and $1$ are both $0$.
I wonder if there are any known properties or uses of this function. Maybe cryptographic, or fast binary multiplication? Or perhaps it's used to prove some number theoretic result?
 A: Yes, this is rather well known, especially to folks in the formal-group racket.
You’re dealing with the formal group $\mathscr M(x,y)=x+y+xy=(1+x)(1+y)-1$, the formal group associated to multiplication, or in fancier (but perhaps rather sloppy) language, the formal group of the multiplicative group scheme $\mathbf G_{\mathrm m}$.
I hope you see that my formula for $\mathscr M$ is just a recoordinatization of multiplication, with the neutral element moved from $1$ to $0$. Your construction of the logarithm, which is a homomorphism from $\mathscr M$ to the additive formal group $\mathscr A(x,y)=x+y$, is perfect. In fact, it can be expressed by the $p$-adically convergent power series, $-\sum_1^\infty(-x)^n/n$, which you know from Calculus, but with the $1+x$ replaced by $x$. A wonderful thing about it is that it remains convergent whenever you substitute for $x$ something that’s closer to $0$ than $1$ is: in technical terms, $\log(a)$ is convergent for all $a$ with $|a|_p<1$, equivalently $v_p(a)>0$. So, for instance, you can plug $\sqrt2$ in for $a$ in your $2$-adic logarithm, and get a good value.
Only one other thing to say for now, and that’s that you don’t need to consider all $z\to0$ in your formula, you can just take powers of $2$ (powers of $p$ in the general case).
