I'm asking for a closed form (an exact value) of the equation solved for $x$

$$\ln (x+1)=\frac{x}{4-x}$$

$0$ is trivial but there is another solution (approximately 2,2...).

I've tried with Lambert's W Function or with Regularized Gamma Functions but I didn't get so far.

  • $\begingroup$ We can have a very good closed form approximation but I don't think that there is a closed form solution. $\endgroup$ – Claude Leibovici Feb 19 '15 at 16:52
  • $\begingroup$ yes $x \approx 2.13242$. but there is no closed form. $\endgroup$ – Arashium Feb 19 '15 at 16:55
  • 1
    $\begingroup$ Define the $Oliveira$ function $Ol(y)$ as the unique non-zero solution of the equation $\ln(x+1)=\frac{x}{y-x}$ (if it exists). Then the closed-form solution is equal to $Ol(4)$. :) $\endgroup$ – Paul Feb 19 '15 at 16:59
  • $\begingroup$ @Paul, that is the point. many closed forms are just conventions. $\endgroup$ – Arashium Feb 19 '15 at 17:02
  • $\begingroup$ Yes but i mean a formula involving well known functions, like Lambert's one. $\endgroup$ – Renato Faraone Feb 19 '15 at 17:07

Burniston and Siewert solved the equation:


Your equation can be reduced to the form:

$$5z-4= e^{z-1}z$$

which can be solved by Burniston Siewert integral

== References ==

[68] C. E. Siewert and E. E. Burniston, "Solutions of the Equation $ze^z=a(z+b)$," Journal of Mathematical Analysis and Applications, 46 (1974) 329-337.


  • $\begingroup$ Thanks, I'll read this paper! $\endgroup$ – Renato Faraone Mar 24 '15 at 9:23

As said in comments, there is no closed form for the solution.

However, we can generate closed forms of very good approximations starting with as estimate $x_0$ written as a rational number and apply one single iteration of Newton method to get $x_1$ given, for the case of this function, by $$x_1=\frac{(16-9 x_0) x_0-(x_0-4)^2 (x_0+1) \log (x_0+1)}{(x_0-12) x_0+12}$$ Using $x_0=\frac {11}5$ as you gave in the post this will lead to $$x_1=\frac{1045+1296 \log \left(\frac{16}{5}\right)}{1195}\approx 2.135935941$$ Using this result, make $x_0=\frac {2136}{1000}$ which leads to $$x_1=\frac{13450125+21281288 \log \left(\frac{392}{125}\right)}{17713875}\approx 2.132426289$$ and you could repeat this process as long as you wish.


I didn't sleep well since the day this problem came into my life so, working really hard, I personally found how to solve this kind of problems with a generalization of Lambert's function, here's my own answer:

$\ln(x+1)={x\over {4-x}}$

let $y=x+1$

$\ln(y)=-{{1-y}\over {5-y}}$

$y=e^{-{{1-y}\over {5-y}}}$

let $z={{1-y}\over {5-y}}$

$e^{-z}={{5z-1}\over {z-1}}$

$\frac 1 5={{z-\frac 1 5}\over {z-1}}e^z$

which is an equation of the form $e^{cx}{{x-a}\over {x-b}}=k$ that has solution(s):

$x=a+\frac 1 cW_{-ke^{-ca}}(cke^{-ca}(a-b))$

so we can now find that $z$ is equal to:

$z=\frac 1 5+W_{-\frac 1 5 e^{-\frac 1 5}}(-\frac 4 {25}e^{-\frac 1 5})$

So the solutions of the equation for $x$ are $0$ and $\frac {4z} {z-1}$


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