sci, how can I find x in this equation? e=x^(x) pic unrelated
>you dont
x^x = e^(log(x^x)) = e^(x·log(x))
=>
log(x) == 1/x
and now you gotta check the productlog
why don't you just see where x^x=e in a graph?
>>7806878
Too inexact
>>7806862
[eqn] e=x^x \implies 1-x \ln (x) = 0 [/eqn] Then use Newton's method and you get that [math] x \approx 1.76 [/math] or if you don't like that then use Lamberts W-function then [eqn] x = e ^{W(1)} [/eqn]