This is my first post here, so I'll try my best.

I want to solve the root for this equation $$x+\log(x-1)\cdot (1-x) = 0$$ I know I require the Lambert $W$ function, but I always get to this form $$(x-1)e^{x-1} = \frac{(x-1)^{x}}{e}$$ $$x = W(\frac{(x-1)^{x}}{e})+1$$ which isn't very helpful. WolframAlpha states that the root is $$x = e^{W(\frac{1}{e})+1}+1$$ May anyone please give me guidance on how to further my work, or tell me where my errors lie? Thank you very much.


Your problem is, in the $W$ function there shouldn't be any variable $x$. We can solve the equation following way. $$x+\log(x-1)\cdot(1-x) = 0 \implies \log(x-1)=\frac{x}{x-1} \implies x-1=e^{\frac{x}{x-1}}$$ Now observe that $\frac{x}{x-1}=\frac{1}{x-1}+1$, so we can divide both sides by $e$, and we get $$ \frac{1}{e}(x-1)=e^{\frac{1}{x-1}} \implies \frac{1}{e}=\frac{1}{x-1}e^{\frac{1}{x-1}} \implies \frac{1}{x-1}=W(\frac{1}{e}) \implies x=\frac{1}{W(\frac{1}{e})}+1 $$

  • $\begingroup$ Thank you so much, Adam! I really appreciate your help, and I didn't expect to receive it so quickly. Have a nice day! $\endgroup$ – tomoki May 26 '18 at 1:56
  • $\begingroup$ Thank you. Always pleasure to help someone. I hope you have a great day too. $\endgroup$ – Jakobian May 26 '18 at 2:01
  • $\begingroup$ Tick correct answer box big boi. Give the man an apple. $\endgroup$ – Tony Hellmuth May 26 '18 at 2:19
  • 1
    $\begingroup$ @TonyHellmuth Sorry, I didn't realize that box was there. I have ticked it. Thanks so much, again. $\endgroup$ – tomoki May 26 '18 at 14:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.