Solutions of $n^{n}=2n$? What is the best way to find solutions to $n^{n}$ and $2n$? Someone suggested the Lambert W function, but I cannot find a way to set it up in a way to use that. Any other suggestions?
Meaning, what are the solutions for $n$ in
$$n^n=2n$$
 A: $n = 2$ is a straighforwardly verifiable solution. 
By inspection of the graph of $y = n^n - 2n$, there's another zero in the vicinity of$~0.34...$ that can be found to arbitrary precision numerically.
By noting that $\frac{d(n^n - 2n)}{dn} = n^n(\ln(n)+1) - 2$, the function has only one global minima for $n > 0$ and is increasing elsewhere. Therefore these are the only two zeros.
A: to make calculation easier...we can start from
$$n^n=2n$$
$$\implies n^{n-1}=2$$
$$\implies n^{n-1}-2=0$$
Now let's assume $$f(n)=n^{n-1}-2$$
Let's find the root using Newton's method of calculating numerical root of an equation.
the rule is,
$$ x_{n+1}=x_n-\dfrac{f(x)}{f'(x)}$$
so,we will need $f'(n)$.So,let's calculate it first.
here,if you calculate the derivative you will find,
$$f'(n)=n^{n-1}\ln 10(\dfrac{n-1}{n \ln 10}+\log n)$$
So,according to Newton's method 
$$n_{r+1}=n_r-\dfrac{n_r^{n_r-1}-2}{n_r^{n_r-1}\ln 10(\dfrac{n_r-1}{n_r \ln 10}+\log n_r)}$$
So,let's start our first assumption 
$$n_o=0.5$$
and continuing the iteration process,we will get
$$0.25535...\\0.319893...\\0.344004...\\0.346305...\\0.3463233...\\0.3463233$$
From here we can conclude that,one root is $n \approx 0.346$
Again if we start the iteration process taking the first assumption $n_o=2.5$
we get,
$$2.17418...\\2.02494...\\2.00055...\\2.0000002...\\2$$
from here we can conclude that another root is $n=2$
Here, $$0.346^{0.346-1}\approx 2$$
again ,$$2^{2-1}=2$$
Hence,yes our calculated roots are correct.
Now,if you want to feel the Lambert W function here, then just do a little work...
$$n^n=2n$$
$$\implies n^{n-1}=2$$
$$\text{so,}~n=W(2)=\Omega$$
Now,we can write, $$\Omega^{\Omega-1}-2=0$$
Now,continue the same process as mentioned before. 
