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.

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. 
