# Tetration Forum

Full Version: simple base conversion formula for tetration
You're currently viewing a stripped down version of our content. View the full version with proper formatting.
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 $c > \eta$