However it never became clear to me what your actual conjecture is as you always work with approximations. So I made some own thoughts:
Let \( \alpha_1 \) be the regular Abel function developed at the primary fixed point \( a \) and let \( \alpha_2 \) be the regular Abel function developed at \( \overline{a} \).
Then we can set \( \alpha(x):=\frac{\alpha_1(x)+\alpha_2(x)}{2} \) and this is again an Abel function (slog) for \( e^x \):
\( \alpha(1)=\frac{\alpha_1(1)+\alpha_2(1)}{2}=0 \).
\( \alpha(e^x)=\frac{\alpha_1(e^x)+\alpha_2(e^x)}{2}
=\frac{\alpha_1(x)+1+\alpha_2(x)+1}{2}=\alpha(x)+1 \).
This would also yield an Abel function for other bases \( b>\eta \) and for any pair of conjugated fixed points.
Is your conjecture now that Andrew's slog is this primary \( \alpha \)? For me it would also be interesting what happens for \( 1<b<\eta \). In continuation of the above idea for a real fixed point we simply would get its regular iteration. However which of the both real fixed points is the happy one for which the regular Abel function is Andrew's slog?
Let \( \alpha_1 \) be the regular Abel function developed at the primary fixed point \( a \) and let \( \alpha_2 \) be the regular Abel function developed at \( \overline{a} \).
Then we can set \( \alpha(x):=\frac{\alpha_1(x)+\alpha_2(x)}{2} \) and this is again an Abel function (slog) for \( e^x \):
\( \alpha(1)=\frac{\alpha_1(1)+\alpha_2(1)}{2}=0 \).
\( \alpha(e^x)=\frac{\alpha_1(e^x)+\alpha_2(e^x)}{2}
=\frac{\alpha_1(x)+1+\alpha_2(x)+1}{2}=\alpha(x)+1 \).
This would also yield an Abel function for other bases \( b>\eta \) and for any pair of conjugated fixed points.
Is your conjecture now that Andrew's slog is this primary \( \alpha \)? For me it would also be interesting what happens for \( 1<b<\eta \). In continuation of the above idea for a real fixed point we simply would get its regular iteration. However which of the both real fixed points is the happy one for which the regular Abel function is Andrew's slog?
