Believe it or not, this isn't homework. It's been many years since grade school, and I'm trying to brush up on these things. But my intuition isn't helping me here.
How can I solve for $n$ in the equation $n \log n = C$?
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1$n\log n = C$ is equivalent to $n^n = e^C$ which has no closed form solution for $n$. There is a solution for $n$ in terms for [Lambert W function](http://en.wikipedia.org/wiki/Lambert%27s_W_function). – 2012-08-30
4 Answers
If what you want is to solve for $n$, there is no simple way. The solution has $n=e^{W(C)} = \frac C{W(C)}$ where $W()$ is the Lambert W function.
Since $n \ln n = \ln n^n$, just raise $e$ to each side and you get $n^n = e^C$.
If you want to solve for $n$, I don't know of any method that will work (usually in Computer Science we use approximations or we solve it numerically).
I will take the base of the logarithm to be $e$. (If you are in a different base; just replace the $e$.)
$n\log n = C$
$\log n = \frac{C}{n}$
$e^{\frac{C}{n}} = n$
$(e^C)^{\frac{1}{n}} = n$
$e^C = n^n$
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0I don't quite see how this solves for $n$. – 2018-03-10
You can show that the function $g(x)=x\ln x$ is negative for $0
It is equal to zero for $x=1$ and is greater than zero for $x>1$.
Therefore we have to make some initial assumptions on $C$. We either have $C\in[-\frac{1}{e},0)$, $C=0$ or $C>0$.
Note that for $C=0$ we have $n=1$ is the solution.
Because $g(x)$ is not one-to-one on $(0,1)$ and has a local minimum we have two solutions for $C\in[-1/e,0)$ so we will disregard this.
There are various ways of solving the equation numerically and here I will just show another idea.
Consider the function $f_C(x)=\frac{C}{\ln x}.$ The fixed point of $f_C(x)$, $x_f$ is the $n$ that solves your equation.
If the fixed point is attracting you have a way of approximating it. To be attracting we need $|f_C'(x_f)|<1$. This is equivalent to
$C
but $C=x_f\ln x_f$ and so if $x_f>e$ we have
$C=x_f\ln x_f
So if the solution to the equation is greater than $e$ the fixed point is attracting. Therefore if we can choose an $x_0$ close enough to $x_f$ then the iterations of $x_0$ under $f_C$ will converge to $x_f$.
For example, suppose I am looking at $400=n\ln n$ and I start with $x_0=3$. I find after eight iterations that I have correct to four significant figures $n=89.09$.