How to express the solution of $x^{x+1}=a$ using Lambert function? I know that the standard Lambert function can be used to describe the solution of $x^x=a$. I wonder if $x^{x+1}=a$ can be addressed similarly.


1 Answer 1


Sorry, my answer is wrong. As it was pointed out by "Simply Beautiful Art" $x\ne W(z)$ but $x=e^{W(z)}.$

It is known that $$W'(z)=\frac{W(z)}{z(1+W(z))}.$$ Let $z=\log t$. Then $x=W(z)$ is a root of the equation $x^x=t$, in particular $x\log x=\log t=z.$ It means that $$W'(z)=\frac{x}{(x+1)\log t}=\frac{1}{(x+1)\log x}=\frac{1}{\log x^{x+1}},\quad x^{x+1}=e^{\frac{1}{W'(z)}}.$$ So solution of the equation $x^{x+1}=a$ is $x=W(z)$, where $z$ is defined by $W'(z)=1/\log a.$

  • $\begingroup$ What about if $x$ is in the denominator, i.e., how to solve $\left(\frac{a}{x}\right)^{x+1}=b$? Thank you $\endgroup$
    – lchen
    Mar 21, 2020 at 14:40
  • $\begingroup$ @lchen It is not clear. $\endgroup$ Mar 21, 2020 at 15:16
  • $\begingroup$ Imean to solve the equation $\left(\frac{a}{x}\right)^{x+1}=b$. $\endgroup$
    – lchen
    Mar 21, 2020 at 15:32
  • $\begingroup$ @lchen I don't know. $\endgroup$ Mar 21, 2020 at 15:36
  • 2
    $\begingroup$ 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$$(W')^{-1}(z)=x\ln(x),~x=\frac e{W^{-1}(\ln(z)+1)}$$ $\endgroup$ Mar 22, 2020 at 0:43

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