How can I solve the equation $n\log_2n = 10^6$? I was solving some exercises and the most of them I found the algebraic properties on the web.
But in the equation $n\log_2 n = 10^6$ I have no idea, I tried several ways to solve the equation and none of them worked.
Thanks!
 A: The solution of this kind of equation has no closed form. But it can be represented by a sort of special function, called Lambert W function. Write original equation as
$$m2^m=10^6$$
by setting $m=2^n$, then the solution is
$$m=W_2(10^6)$$
Or, 
$$n=\log_2(W_2(10^6))=\frac{10^6\ln2}{W_e(10^6\ln2)}$$
A: Rearrange: $\log_2{n^n} = 10^6$
$n^n = 2^{10^{6}} = 2^{1000000}$ 
This equation can't be solved with elementary functions, you need to use the Lambert W function.
Approximating: $n \approx 62746.126$ 
A: As said in previous answers and comments, equations such as  $$n\log_a n = b$$ have no elemental solutions. Only Lambert function provides a solution which is (for the algebraic case)
$$n=\frac{b \log (a)}{W(b \log (a))}$$ For large values of $x$, $$W(x) \simeq \log (x)-\log (\log (x))+\frac{\log (\log (x))}{\log (x)}$$ So, in your case where $x=10^6 \log(2)$, this would give an estimate equal to $62766.1$ while the exact solution is $62746.1$.  
By the way, any equation of the form $$a+b x+c \log (d x+e)=0$$ has solutions which can be expressed using Lambert function $$x=\frac{c d W\left(\frac{b e^{\frac{b e}{c d}-\frac{a}{c}}}{c d}\right)-b e}{b d}$$
