Well I finally decided to look a little deeper at my suggested base conversion, and to my surprise, the formula just popped out at me.

given the definition \( k \ge 0 \):

\( a\,\,\bigtriangleup_{-k}^e\,\, b = \ln^{\circ k}(\exp^{\circ k}(a) + \exp^{\circ k}(b)) \)

we see instantly:

\( e^x\,\,\bigtriangleup_{-k}^e\,\,e^y = e^{x\,\,\bigtriangleup_{-k-1}^e\,\,y} \)

which works for k = 0 as well since \( \bigtriangleup_{1}^e \) is multiplication.

It's easy now to generate a formula which works for base conversion using these operators. The notation for such is very cumbersome however, therefore I'll write it out step by step for 2, 3, 4.

\( b^b = e^{\ln(b) \cdot e^{\ln(b)}} = e^{e^{\ln(b) + \ln(\ln(b))}} = \,\,^{2+\text{slog}_e(\ln(b) + \ln(\ln(b)))}e \)

\( b^{b^b} = e^{\ln(b) \cdot e^{\ln(b) \cdot e^{\ln(b)}}} = e^{e^{e^{\ln(b) + \ln(\ln(b))\, \bigtriangleup_{-1}^e \,\ln(\ln(\ln(b)))}}}= \,\,^{3+\text{slog}_e(\ln(b) + \ln(\ln(b)) \bigtriangleup_{-1}^e \ln(\ln(\ln(b))))}e \)

and which I think the process becomes obvious by this point

\( b^{b^{b^{b}}} = \,\,^{4 + \text{slog}_e(\ln(b) + \ln(\ln(b)) \,\bigtriangleup_{-1}^e\, \ln(\ln(\ln(b)))\,\bigtriangleup_{-2}^e\,\ln(\ln(\ln(\ln(b)))))} e \)

and which by generality becomes:

\( ^k b = \,\,^{k + \text{slog}_e(\ln(b) + \ln^{\circ 2}(b) \,\bigtriangleup_{-1}^e\,\ln^{\circ 3}(b)\,\bigtriangleup_{-2}^e\,\ln^{\circ 4}(b)\,...\,\bigtriangleup_{-k+1}\,\ln^{\circ k-1}(b)\,\bigtriangleup_{-k+2}^e\,\ln^{\circ k}(b))}e \)

where we are sure to evaluate the highest operator first.

this formula is very easily generalized to include conversion from base \( b > \eta \) to any base \( c > \eta \)

given the definition \( k \ge 0 \):

\( a\,\,\bigtriangleup_{-k}^e\,\, b = \ln^{\circ k}(\exp^{\circ k}(a) + \exp^{\circ k}(b)) \)

we see instantly:

\( e^x\,\,\bigtriangleup_{-k}^e\,\,e^y = e^{x\,\,\bigtriangleup_{-k-1}^e\,\,y} \)

which works for k = 0 as well since \( \bigtriangleup_{1}^e \) is multiplication.

It's easy now to generate a formula which works for base conversion using these operators. The notation for such is very cumbersome however, therefore I'll write it out step by step for 2, 3, 4.

\( b^b = e^{\ln(b) \cdot e^{\ln(b)}} = e^{e^{\ln(b) + \ln(\ln(b))}} = \,\,^{2+\text{slog}_e(\ln(b) + \ln(\ln(b)))}e \)

\( b^{b^b} = e^{\ln(b) \cdot e^{\ln(b) \cdot e^{\ln(b)}}} = e^{e^{e^{\ln(b) + \ln(\ln(b))\, \bigtriangleup_{-1}^e \,\ln(\ln(\ln(b)))}}}= \,\,^{3+\text{slog}_e(\ln(b) + \ln(\ln(b)) \bigtriangleup_{-1}^e \ln(\ln(\ln(b))))}e \)

and which I think the process becomes obvious by this point

\( b^{b^{b^{b}}} = \,\,^{4 + \text{slog}_e(\ln(b) + \ln(\ln(b)) \,\bigtriangleup_{-1}^e\, \ln(\ln(\ln(b)))\,\bigtriangleup_{-2}^e\,\ln(\ln(\ln(\ln(b)))))} e \)

and which by generality becomes:

\( ^k b = \,\,^{k + \text{slog}_e(\ln(b) + \ln^{\circ 2}(b) \,\bigtriangleup_{-1}^e\,\ln^{\circ 3}(b)\,\bigtriangleup_{-2}^e\,\ln^{\circ 4}(b)\,...\,\bigtriangleup_{-k+1}\,\ln^{\circ k-1}(b)\,\bigtriangleup_{-k+2}^e\,\ln^{\circ k}(b))}e \)

where we are sure to evaluate the highest operator first.

this formula is very easily generalized to include conversion from base \( b > \eta \) to any base \( c > \eta \)