Is it possible to solve $k = \frac{x}{\ln(x)}$ for $x$? Is it possible to solve $k = \frac{x}{\ln(x)}$ for $x$? My suspicion after a fruitless hour of manipulation is that it is not.
 A: It can be solved with the Lambert W function.
First let $y=\frac1x$. 
Then let $z=\ln(y)$, and use the Lambert W function. Then substitute back. 
\begin{align*} k &= \frac{x}{\ln x} \\  k &= \frac{\frac1y}{\ln \frac1y} \\  k &= \frac{1}{y\ln \frac1y} \\ k &= -\frac{1}{y\ln y} \\ -\tfrac{1}k &= y\ln y 
\\ -\tfrac{1}k &= e^zz
\\ W\left(-\tfrac{1}k\right) &= z
\\ W\left(-\tfrac{1}k\right) &= \ln(y)
\\ e^{W\left(-\tfrac{1}k\right)} &= y
\\ e^{W\left(-\tfrac{1}k\right)} &= \tfrac1x
\\ e^{-W\left(-\tfrac{1}k\right)} &= x
\end{align*}
Hence $x= e^{-W\left(-\tfrac{1}k\right)} $. 
Note that the Lambert W function is only defined for $x\geq-\tfrac{1}e$. Therefore we need $k>e$ or $k<0$. Otherwise there is no solution. 
When $k>e$, we need to consider both branches of W. Then $x= e^{-W_0\left(-\tfrac{1}k\right)} \vee e^{-W_{-1}\left(-\tfrac{1}k\right)}$. 
Note that we need the Lambert W function. If $e^{-W\left(-\tfrac{1}k\right)}$ could be written with just elementary functions, then $W(x)$ can be written using just elementary functions. But it has been proven that it can't. 
A: Consider the function $f(x)=x-k\ln x$, defined for $x>0$. We have
$$
\lim_{x\to\infty}f(x)=\infty
$$
and
$$
\lim_{x\to0}f(x)=
\begin{cases}
\infty & \text{if $k>0$}\\[4px]
-\infty & \text{if $k<0$}
\end{cases}
$$
We already see that a solution for $f(x)=0$ (which is the same as your equation) exists when $k<0$, but this is not sufficient.
Consider the derivative
$$
f'(x)=1-\frac{k}{x}=\frac{x-k}{x}
$$
Case $k<0$
If $k<0$, the derivative is positive everywhere, so the function is increasing and we get a single solution.
Case $k=0$
There is no solution, because this would imply $x=0$, but the equation makes no sense for $x\le0$.
Case $k>0$
So let $k>0$. We see that the function has a minimum at $k$, with
$$
f(k)=k(1-\ln k)
$$
The number of solutions then depends on whether $1-\ln k$ is positive (no solution), zero (one solution) or negative (two solutions). Clearly, $1-\ln k>0$ if and only if $0<k<e$.
Summary


*

*no solution if $0<k<e$;

*one solution if $k=e$ or $k<0$

*two solutions if $k>e$.
With numerical methods you can approximate the solutions to the required accuracy.
