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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.

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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 $$ \frac{1}{e}(x-1)=e^{\frac{1}{x-1}} \implies \frac{1}{e}=\frac{1}{x-1}e^{\frac{1}{x-1}} \implies \frac{1}{x-1}=W(\frac{1}{e}) \implies x=\frac{1}{W(\frac{1}{e})}+1 $$

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  • $\begingroup$ 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
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    $\begingroup$ @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

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