Ahhh. Thank you. That makes a lot of sense. I assume b is a constant, then I solve for b as a function of x! I am pretty dense.
Anyways, here's something else I found that may be of use,
Let's say that \( f(x)=e^{g(x)} \).
Then by definition
\( f(f(x))=e^{g(e^{g(x)})}=e^x\\\Rightarrow g(e^{g(x)})=x\\\Rightarrow e^{g(x)}=g^{-1}(x) \)
Now substitute g(x) for x,
\( e^{g(g(x))}=g^{-1}(g(x))=x\\\Rightarrow g(g(x))=ln(x) \)
So, in conclusion, the half-iterate of e^x is the exponential of the half-iterate of ln(x).
Not sure if that is useful but I think it's kinda neat. Hopefully I didn't make another mistake, but it's always possible.
Anyways, here's something else I found that may be of use,
Let's say that \( f(x)=e^{g(x)} \).
Then by definition
\( f(f(x))=e^{g(e^{g(x)})}=e^x\\\Rightarrow g(e^{g(x)})=x\\\Rightarrow e^{g(x)}=g^{-1}(x) \)
Now substitute g(x) for x,
\( e^{g(g(x))}=g^{-1}(g(x))=x\\\Rightarrow g(g(x))=ln(x) \)
So, in conclusion, the half-iterate of e^x is the exponential of the half-iterate of ln(x).
Not sure if that is useful but I think it's kinda neat. Hopefully I didn't make another mistake, but it's always possible.

