Find the equation of the tangent line to the curve y=ln(lnx) which is perpendicular to the line ex+y =0

So I took the derivative and got 1/(xlnx)

I know that the slope of the normal line is 1/e so i substituted that in for y' and tried solving for x. this is where I was stuck

1/e = 1/(xlnx)

xlnx = e