I'm playing with math I only partially understand, which means I know enough to be dangerous. Why doesn't this transformation work? I wasn't necessarily expecting it to but I also don't understand where a mistake was made.
First, take an equation with real valued variables and coefficients:
$$x + y = 2 \ b$$
$$e^{\log(x)} + e^{\log(y)} = 2 \ e^{\log(b)}$$
Now extend the equation in the complex plane by introducing a new parameter $s \in \mathbb{C}$ such that the equation is analytic in the open unit disk (at least) and the analytic continuation of the function at $s=1$ gives us the original equation:
$$e^{s \log(x)} + e^{s \log(y)} = 2 \ e^{s \log(b)}$$
Now break the exponents into infinite series:
$$\sum_{n=0}^{\infty} s^n \frac{\log(x)^n}{n!} + \sum_{n=0}^{\infty} s^n \frac{\log(y)^n}{n!} = 2 \sum_{n=0}^{\infty} s^n \frac{\log(b)^n}{n!}$$
$$\sum_{n=0}^{\infty} \frac{s^n}{n!} \left( \log(x)^n + \log(y)^n\right) = 2 \sum_{n=0}^{\infty} \frac{s^n}{n!} \log(b)^n$$
The equality should hold for all $s$, so equate the various terms of the infinite series:
$$\log(x)^n + \log(y)^n = 2 \log(b)^n$$
This holds for all values of $x, y, b$ at $n=0$. So far so good. However at larger values of $n$ this breaks down. For instance, there's a counter example if $x = e^2, y = e^4$. First solve for $b$ at $n=1$: $$\log(e^2) + \log(e^4) = 2 \log(b)$$ $$2 + 4 = 2 \log(b)$$ $$b = e^3$$
but then: $$\log(e^2)^4 + \log(e^4)^4 = 2 \log(e^3)^4$$ $$2^4 + 4^4 = 2 \cdot 3^4 $$ $$16 + 256 = 2 \cdot 81 $$ $$272 = 162$$
Which isn't true. So I made at least one mistake somewhere. What is the mistake? I assume it's in how I set up the extension into the complex plane?