Solve for x? x*e^x=1.

taking log base e on both sides

logx + log(e^x)=log1
logx+x=0
-x=logx
taking antilog

x=-e^x

x+e^x=0

solve it via newton's raphson's law

x=-0.56714329

Using the Lambert W-Function, the solution is just:
x = W(1).

Note that the natural logarithm function and the Lambert W-Function are very similar. The natural logarithm function is the inverse of e^x; the Lambert W-Function is the inverse of x*e^x.

Alternatively, you can approximate the solution to x*e^x = 1 via Newton's Method. There is no way to get an exact value for x using elementary functions (note that the Lambert W-Function is not elementary).

I hope this helps!

Use Newton method of roots
If r is an approximation to f(r)=0, then a better approximation is given by r(better)=r- f(r)/f'(r)
y= x*e^x - 1
y' = x*e^x + e^x
let r=.5
Next r = .5 - (.5*e^.5 - 1)/(.5*e^.5 + e^.5) = .571
Plug .571 in above equation and get .56716
Ultimate solution is r=.56714329040978

plot the curves e^x and 1/x to find the place of the root.

find the root of f(x)=1-xe^x by any calculator or software.

Try to graph y= x e^x and y=1 and find the intersection point.

They intersect at x=0.5671

0.567