What is the inverse fun of #y=x^x#?

Given:

#y = x^x#

This question is essentially asking us to solve for #x# in terms of #y#.

This cannot be done with elementary functions, but we can use a family of functions called the Lambert #W# function, that provide the inverse to:

#w = ze^z#

namely:

#z = W_n(w)#

where #n in ZZ# is the index of the branch, with #W_0# being the principal branch, providing the unique real valued solution when #w > 0#.

Putting #x = e^t# in our given equation we have:

#y = x^x = (e^t)^(e^t) = e^(t e^t)#

So taking logs, we have:

#ln y = t e^t#

or more generally for complex solutions:

#ln y + 2kpi i = t e^t" "# for any #k in ZZ#

So using the Lambert #W# function, we have:

#t = W_n (ln y)#

or more generally

#t = W_n (ln y + 2kpi i)#

Then taking the exponent of both sides:

#x = e^t = e^(W_n(ln y))#

or more generally:

#x = e^t = e^(W_n(ln y + 2kpi i))#

So the inverse of:

#f(x) = x^x#

is:

#f^(-1)(y) = e^(W_n(ln y))#

or more generally:

#f^(-1)(y) = e^(W_n(ln y + 2kpi i))#

or if you prefer in terms of #x# and #y#, swap them in the expression we found to get:

#y = e^(W_n(ln x))#

or more generally:

#y = e^(W_n(ln x + 2kpi i))#

This is actually a family of functions indexed by branch number #n# rather than a single function.

For #x > 0# we can choose the real valued principal branch and write:

#y = e^(W_0(ln x))#