x^(x*x^x) = 2

(x^x)^(x^x) = 2

k = x^x

k^k = 2

k*ln(k) = ln(2) → Log of both sides

ln(k) * e^ln(k) = ln(2) → k = e^ln(k)

f(ln(k)) = ln(2)

ln(k) = W(ln(2))

ln(x^x) = W(ln(2))

ln(x)*e^ln(x) = W(ln(2)) → Same step as noted earlier

f(ln(x)) = W(ln(2))

ln(x) = W(W(ln(2))

x = e^W(W(ln(2)))

x ≈ 1.3799703966 (via Wolfram|Alpha, seems to be the correct value)