What is $Z$ such that $e^X e^Y e^{-X}=e^Z $?

Question

We know that $e^Z = e^X e^Y$ can be solved using Dynkin's formula.

I am thinking of how to similarly find $Z$ for the product in title. Do not need the actual answer, a hint suffices!

Answer 1

By the Baker–Campbell–Hausdorff formula we have $$e^{X} e^{Y} e^{-X}= e^{Y+\left[X,Y\right]+\frac{1}{2!}[X,[X,Y]]+\frac{1}{3!}[X,[X,[X,Y]]]+\cdots}$$ The exponent $Z$ on the right hand side can be written as $$Z = \sum_{n=0}^\infty\frac{a_n}{n!}~~\text{where}~~a_{n+1} = [X,a_n]~~\text{with}~~a_0 = Y$$ For the special case where $X$ and $Y$ commute then this simplifies to $e^Y = e^Z$. As pointed out by John below in general $Z=Y$ is not the only solution as the matrix logarithm is not unique.