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When trying to solve the equation in the title with WA, it produced the following as the solution:

enter image description here

now, if you divide the numerator and denominator by $y$ and set $z:=-\frac{a}{y}$ the solution becomes
$$\frac{a}{W_n(z\,e^z)-z}\ =\ \frac{a}{z-z}\ =\ \frac{a}{0}\ =\ \infty$$

Questions:

  • can $x\cdot(e^{\frac{a}{x}}-1)-y=0$ be solved for $x$ and can the solution be expressed with the Lambert $W_n$ function resp.
  • can the correctness of the result generated by by WA be explained?
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  • $\begingroup$ Did you try asking Stephen Wolfram stephenwolfram.com? $\endgroup$
    – markvs
    Apr 3, 2022 at 16:59
  • $\begingroup$ @markvs not yet... $\endgroup$ Apr 3, 2022 at 17:08

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Define $x'=x/a$, $y'=y/a$, then the real-valued solution to $x'(e^{1/x'}-1)=y'$ is $$x'=\frac{-y'}{1+y'W_0(-e^{-1/y'}/y')}\;\;\text{for}\;\;0<y'<1,$$ $$x'=\frac{-y'}{1+y'W_{-1}(-e^{-1/y'}/y')}\;\;\text{for}\;\;y'>1.$$ There is no solution for $y'<0$.
A numerical test: for $y'=1/2$ this gives $x'=-0.6275$ and for $y'=2$ this gives $x'=0.795905$, which indeed solves the equation.


The "$\infty$" result at the end of the OP appears because the identification $W_n(ze^z)=z$ is mistaken. (The correct equation is $W_n(z)e^{W_n(z)}=z$.)

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    $\begingroup$ Note $W_n$ with subscripts other than $0,-1$ give us complex solutions. The OP did not specify real when asking Wolfram Alpha, so it assumed complex. $\endgroup$ Apr 3, 2022 at 21:16
  • $\begingroup$ indeed, I will remove that line, thanks. $\endgroup$ Apr 3, 2022 at 21:19

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