Solving x^y=y^x

[Albert Chan]

(10-24-2018 07:37 PM)Dieter Wrote:  The WP34s, for example, handles this easily:

LN
RCL/ L
CHS
Wp
RCL/ L

Enter x, get y and vice versa.

Thanks, Dieter. I was curious how above get derived ...

Found a very close solution for infinite tetration c = x^c, or ln(c)/c = ln(x).
From the video (~ 7:00), using basic algebra, he got c = W(-ln(x)) / (-ln(x))

For x^y = y^x, ln(y)/y = ln(x)/x, so just substitute above: c=y, ln(x)=ln(x)/x
--> y = W(-ln(x)/x) / (-ln(x)/x)