how to get value of n if the value of n * log n is given? 
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How can I solve for $n$ in the equation $n \log n = C$? 

how to get value of n if the value of n * log n is given ?
I am stuck with this:
n log n = 6 * 10^6

how to tackle such equations ?
 A: If you can't solve it by inspection, it is unlikely there is a "nice" solution.  In that case you can use the fact that $\log n \ll n$ and varies slowly.  So in this problem, you can write it as $n=\frac {6 \cdot 10^6}{\log n}$  Then for a first guess use $n=6 \cdot 10^6$, take the log to get $6.778$, plug that in and evaluate $n$, iterating until it converges.
A: In general, the equation $x \log(x) = a$ cannot be solved for $x$ in closed form in terms of elementary functions. "Elementary functions" refers to the functions formed via the composition of the basic arithmetic operations +, -, *, /, the exponential and logarithm functions, and the trigonometric functions (though the inclusion of these last functions is redundant if one uses complex numbers, due to Euler's formula.), with the composition being finite.
Usually, elementary solutions are considered the best, though, as this shows, they are in many cases not possible. (There are analogies of this in other areas of mathematics -- e.g. in Euclidean geometry, constructions relying only on compass and straightedge are considered the best. Again, there are many cases in which such construction is not possible.) This is because elementary functions are simple and have numerous useful properties.
That being said, if one is willing to venture outside the realm of elementary functions, one can solve this equation. One commonly used non-elementary, but still simple to describe, function is the Lambert W function, defined as the inverse of $f(x) = xe^x$. Note this is similar to a logarithm function, but not quite the same. Such an inverse is not a proper function, since $f(x) = xe^x$ is not injective, however, we usually define a "principal branch" similar to, say, square root, by inverting on the interval $[-1, \infty)$. The Lambert W-function is denoted $W(x)$. To use it to solve your equation, make the substitution $x = e^u$. EDIT: In fact, this can be used as a proof that there is no elementary solution, because this shows that if you can solve the equation, you can express the Lambert W function in terms of its solution, but we know that one is not elementary. (The proof of that is much harder, however.)
However, if your $\log$ is base-10, then there is an elementary solution, namely $n = 10^6$! The form of the right-hand side makes me wonder whether a base-10 log was meant.
A: For this kind of problem, with a specific number given, an iteration formula is best. What matters most is that n appear once in  a direct way (the factor n) and once in a slowly-varying way (the logarithm).   
(I'll take the log to be natural base.)
Since log(n) varies slowly, and is "small" relative to n, we'll take it to be "constant" and set it to 1 just for a moment.  Then n = the given number, 6E6.  Obviously wrong; log(6E6)  (using base e) is about 15.  Okay, take n to be 6E6/15.607   and plug that into the logarithm to get a "better" number than 15.607.   Repeat...
n = 6E6
n ← (6.0E6)/log(n)   (repeat)

This converges quickly to n=460147.3...
If you need some sort of analytic formula, exp() both sides and maybe the Lambert W function or something similar will be helpful.  There won't be any solution in terms of common everyday arithmetic and familiar functions.
