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Previously, I solved the special transcendental equation $x=e^{t/\ln(x)}$. The solution is: $x=e^{-\sqrt{t}}$ for $0<x<1.$

One can define an equation: $e^{s/\ln(x)}=e^{t/\ln(1-x)},$ for $s,t \in \Bbb R ,x \in \Bbb R(0,1) $ and $x \ne 0,1. $

Fixing $s,t$ and letting $s=t$ yields a trivial solution $x=1/2.$ However, to solve this equation with $s=1$ and $t=4$ requires solving the polynomial: $(1-x)^4+x-1=0.$ If $s=1$ and $t=5$ it becomes impossible to find a closed form solution.

I found that setting $s=1$ and $t=2,$ yields $x=1/\phi,$ where $\phi$ is the golden ratio. When $s=2$ and $t=4$ this yields the solution $x=\phi-1=1/\phi$. Also, when $s=1/2$ and $t=1/4$, $x=1/\phi^2.$ When $s=1/4$ and $t=1/8,$ $x=1/\phi^2.$ When $s=1/8$ and $t=1/16,$ $x=1/\phi^2.$

Could I express more solutions to this equation for different values of $s,t$ in terms of $\phi$? How many?

J. Zimmerman
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  • Actually your first equation is satisfied by $x = \exp(\pm \sqrt{t})$ (with $t > 0$ if you want real solutions). – Robert Israel Sep 27 '18 at 03:19
  • $e^{s/\ln(x)} = e^{t/\ln(1-x)}$ is equivalent to $e^{s \ln(1-x)} = e^{t \ln x}$, i.e. $(1-x)^s = x^t$. In the case $s=1$, $t=5$ the real solution is $$x = -\frac{1}{3} + \frac{(100+12\sqrt{69})^{1/3}}{6} + \frac{2}{3 (100 + 12 \sqrt{69})^{1/3}}$$ – Robert Israel Sep 27 '18 at 03:34
  • This is because the polynomial $1 - x - x^5 = (1 - x + x^2)(1 - x^2 - x^3)$ is reducible. – Robert Israel Sep 27 '18 at 03:36

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As @RobertIsrael commented, in th mest general case, you look for the zero of function $$f(x)=(1-x)^s - x^t$$ which is the inverse of $$k=\frac s t=\frac{\log(1-x)}{\log(x)}$$ which us eaxctly what I aked in this question.

For $k<1$, I proposed as a first estimate $$x_0\sim\frac 1 {1+k^{-\log_2 (\phi )}}$$ which, for $k=\frac 15$ would give $x_0=0.246505$ while the "exact" solution is $0.245122$.

Using one single iteration of any Newton-lik mathod of order $n$ would give an explicit formula which will be better and better.

Claude Leibovici
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