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Timeline for A variant of Lambert function

Current License: CC BY-SA 4.0

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Mar 22, 2020 at 9:13 comment added lchen Thank you both.
Mar 22, 2020 at 4:53 history edited Alexey Ustinov CC BY-SA 4.0
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Mar 22, 2020 at 4:49 comment added Alexey Ustinov @SimplyBeautifulArt You are write, my answer is wrong.
Mar 22, 2020 at 0:43 comment added Simply Beautiful Art Assuming I haven't made a mistake, using this approach but modified with $\ln(x)=W(z)$ will give us $$W'(z)=\frac1{x(\ln(x)+1)}$$which is reduces to$$\frac e{ex\ln(ex)}$$which is invertible using the Lambert W function. It does go to show, however, that the inverse of $W'$ is solvable in terms of $W$ and is given by$$(W')^{-1}(z)=x\ln(x),~x=\frac e{W^{-1}(\ln(z)+1)}$$
Mar 22, 2020 at 0:19 comment added Simply Beautiful Art Correct me if I'm wrong, but if $z=\ln(t)$ and $x=W(z)$, then $z=xe^x$ and $t=e^z=e^{xe^x}\ne x^x$.
Mar 21, 2020 at 23:37 comment added Simply Beautiful Art Huh I've never thought to use the derivative of the Lambert W function, interesting approach.
Mar 21, 2020 at 15:48 comment added Alexey Ustinov @lchen If you'll replace $x$ by $-x$ and $a$ by $-a$, then you'll get $\left(\frac{x}{a}\right)^{x-1}=b.$
Mar 21, 2020 at 15:44 comment added lchen Thank you all the same. I thought that it was related to Lambert function but could not figure out the solution.
Mar 21, 2020 at 15:36 comment added Alexey Ustinov @lchen I don't know.
Mar 21, 2020 at 15:32 comment added lchen Imean to solve the equation $\left(\frac{a}{x}\right)^{x+1}=b$.
Mar 21, 2020 at 15:16 comment added Alexey Ustinov @lchen It is not clear.
Mar 21, 2020 at 14:40 comment added lchen What about if $x$ is in the denominator, i.e., how to solve $\left(\frac{a}{x}\right)^{x+1}=b$? Thank you
Mar 21, 2020 at 12:37 history answered Alexey Ustinov CC BY-SA 4.0