You can get analytical solutions with Reduce, if you use the definition equation for ProductLog z = w*E^w.
Here calculated for all three equations together with tests.
Reduce[z == w E^w, w]
(* *)
red1 = Reduce[{w/z == E, z + Log[x] == 0, z == w E^w}, x]
rule1 = (Rule @@ First@Cases[#, x == xsol__] &) /@ red1
ProductLog[0, -Log[x]]/-Log[x] - E /. rule1[[1]] // N
ProductLog[-1, -Log[x]]/-Log[x] - E /. rule1[[2]] // N
ProductLog[1, -Log[x]]/-Log[x] - E /. rule1[[3]] // N
(* *)
red2 = Reduce[{w/z == Pi, z + Log[x] == 0, z == w E^w}, x]
rule2 = (Rule @@ First@Cases[#, x == xsol__] &) /@ red2
ProductLog[-1, -Log[x]]/-Log[x] - Pi /. rule2[[1]] // N
ProductLog[1, -Log[x]]/-Log[x] - Pi /. rule2[[2]] // N
ProductLog[-1, -Log[x]]/-Log[x] - Pi /. rule2[[3]] // N
(* *)
red3 = Reduce[{w/z == I, z + Log[x] == 0, z == w E^w}, x]
rule3 = Rule @@ First@Cases[red3 /. C[1] -> c1, x == xsol__]
Table[ProductLog[c1, -Log[x]]/-Log[x] - I /. rule3, {c1, 1, 5}] //
N // Chop
ProductLog[-1, -Log[x]]/-Log[x] - I /. rule3 /. c1 -> 0 // N
Solutions for the three equations
(x -> E^(1/E)) || (x -> E^((I (2 \[Pi] - I))/E)) || (x ->
E^(-((I (2 \[Pi] + I))/E)))
(x -> E^((I (2 \[Pi] - I Log[\[Pi]]))/\[Pi])) || (x ->
E^(-((I (2 \[Pi] + I Log[\[Pi]]))/\[Pi]))) || (x -> \[Pi]^(1/\[Pi])
)
x -> E^(-(1/2) (-1 + 4 c1) E^(2 I c1 \[Pi]) \[Pi])
Solve[ ProductLog[Log[x]] == k Log[x], x]
yields some answers....BTW, questions about W|A are off-topic, except for how to call W|A from within Mathematica. community.wolfram.com is a site for W|A questions. $\endgroup$