Just want to clarify the relation between the decremented exponential \( \operatorname{dexp}(x):=e^x-1 \) and \( \exp_\eta \) with respect to base change by giving the formula. We know that they are affine conjugates:
\( \operatorname{dexp} = \tau^{-1}\circ \exp_\eta \circ \tau \) where \( \tau(x)=(x+1)e \).
We also have \( \log(\log(\exp_\eta(\exp_\eta(x))))=\log(e^{-1}\exp_\eta(x))=-1+e^{-1}x=\tau^{-1}(x) \) thatswhy:
\( \log^{[n]}\circ\exp_\eta^{[n]}=\log^{[n-2]}\circ\tau^{-1}\circ \exp_\eta^{[n-2]}\circ \tau\circ \tau^{-1}=\log^{[n-2]}\circ(\tau^{-1}\circ \exp_\eta\circ \tau)^{[n-2]}\circ \tau^{-1} \)
So this is the relation
\( \fbox{\log^{[n]}\circ\exp_\eta^{[n]} =\log^{[n-2]}\circ\operatorname{dexp}^{[n-2]}\circ \tau^{-1}} \) where \( \tau^{-1}(x)=x/e-1 \).
Update: This formula can be generalized to a base change from \( b=a^{1/a} \) to \( a \), i.e. from a base to one of its fixed points:
\( \fbox{\log_a^{[n]}\circ\exp_b^{[n]} =\log_a^{[n-2]}\circ\operatorname{dexp}_a^{[n-2]}\circ \tau^{-1}} \) where \( \tau^{-1}(x)=x/a-1 \) and \( \operatorname{dexp}_a(x)=a^x-1 \).
\( \operatorname{dexp} = \tau^{-1}\circ \exp_\eta \circ \tau \) where \( \tau(x)=(x+1)e \).
We also have \( \log(\log(\exp_\eta(\exp_\eta(x))))=\log(e^{-1}\exp_\eta(x))=-1+e^{-1}x=\tau^{-1}(x) \) thatswhy:
\( \log^{[n]}\circ\exp_\eta^{[n]}=\log^{[n-2]}\circ\tau^{-1}\circ \exp_\eta^{[n-2]}\circ \tau\circ \tau^{-1}=\log^{[n-2]}\circ(\tau^{-1}\circ \exp_\eta\circ \tau)^{[n-2]}\circ \tau^{-1} \)
So this is the relation
\( \fbox{\log^{[n]}\circ\exp_\eta^{[n]} =\log^{[n-2]}\circ\operatorname{dexp}^{[n-2]}\circ \tau^{-1}} \) where \( \tau^{-1}(x)=x/e-1 \).
Update: This formula can be generalized to a base change from \( b=a^{1/a} \) to \( a \), i.e. from a base to one of its fixed points:
\( \fbox{\log_a^{[n]}\circ\exp_b^{[n]} =\log_a^{[n-2]}\circ\operatorname{dexp}_a^{[n-2]}\circ \tau^{-1}} \) where \( \tau^{-1}(x)=x/a-1 \) and \( \operatorname{dexp}_a(x)=a^x-1 \).