Solving x^y=y^x

[ijabbott]

(10-24-2018 09:58 PM)Albert Chan Wrote:  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)

I like his series of videos on geometric algebra - something that isn't currently handled very well by our favourite, hand-held, symbolic calculators.