Is it possible to solve $k = \frac{x}{\ln(x)}$ for $x$?

Question

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.

Answer 1

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.

Answer 2

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.