x=y^y solve for y

Question

i need an equation where i can input x and find what number when raised to itself will equal that number. i am stuck on how to free the y from log(x)=ylog(y) in the log. i believe you have to use a Lambert function but i cant find information on it online to my level of mathematics( Intro to Calc.)

Answer 1

x = y^y

 

ln x = y ln y

 

ln y = 1/y ln x

 

exp(ln y) = exp(1/y ln x)

 

y = exp(ln x) ^1/y

 

y = x ^1/y

 

dy/dx = 1/y dx/dx

 

dy/dx = 1/y

 

y = 1 / dy/x

 

y = x ^dy/dx

 

ln y = dy/dx lnx

 

(1/ln y) dy = ln x dx

 

∫ dy/ln y = ∫ ln x dx + Const

 

∫ ln x dx = x ln x - x

 

∫ dy/ln y = Σ yⁿ / (n+1)! n = 0 to infinity

 

Hopefully that will help.