How can I solve $1/x = \ln x$? Another way to look at it is  $x^x  =  e$.
It seems possible, but I really don't know.  It seems very helpful to know the exact value where a function's y value is the same as its slope.
 A: Per the comment of clathratus, the Lambert W function is relevant here; since $x\ln x=1$, take $x=1/W(1)$ for some branch $W$ of the function. However, if you actually want a few decimal places of $x$ in a hurry, numerical methods such as the Newton-Raphson method are prudent. Define $f(x):=x\ln x-1$ so $f'=\ln x+1$ and $g(x):=x-\frac{f(x)}{f'(x)}=\frac{x+1}{\ln x+1}$; then we can use the recursion $x_{n+1}=g(x_n)$ (which is probably what ab123 did), provided we choose a sensible seed. The choice $x_0$ should do.
A: green line is a $y=e$ and blue line is graphic for function $x^x$

A: There are several (actually infinitely many) complex solutions to the equation, but there is only one real solution, written as $x=e^{\operatorname{W}(1)}$, where $\operatorname{W}(z)$ is the Product Log Function. Here's a link to more info on this function: http://mathworld.wolfram.com/LambertW-Function.html
A: Hint:
Use the iteration method as follow
$$x_{i+1}=e^{\frac{1}{x_{i}}}$$
start with $x_0=1$ 
After $40$ iterations the value of $x$ will be $1.763222834$
A: You may also use numerical integration: the wanted solution is $\frac{1}{W(1)}$ with $W$ being Lambert's function, hence by the Lagrange inversion theorem
$$ x = \left[\left(\int_{-\infty}^{+\infty}\frac{dt}{\pi^2+(e^t-t)^2}\right)^{-1}-1\right]^{-1}. $$
The integrand function behaves like the normal distribution close to the origin, like the Laplace distribution over $t\gg 0$ and like the Cauchy distribution over $t\ll 0$. This leads to $x\approx 1.76322$.
A: This can be solved either by using Lambert W function or Newton Raphson method.
The above equation can be written as -> 1 = x*ln(x)
1. (Using Lambert W function):
W(x*ln(x)) = W(1)     ---- [1]
as per Lambert W function: W(x*ln(y)) = ln(y)
hence,
ln(x) = W(1)    {substituting in [1]}
so, 
x = e^(W(1))
but W(1) is approximately equal 0.56714329
refer: https://en.wikipedia.org/wiki/Lambert_W_function;
substituting the value (W(1)) above:
x = 1.76322 (approximate value)
2. (Newton Raphson Method):
consider f(x) = x*ln(x) - 1 & solve further using Newton Raphson Method formula for different values of x, starting with 0.
https://en.wikipedia.org/wiki/Newton%27s_method
Hope this helps!
