# A variant of Lambert function

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.

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 $$W'(z)=1/\log a.$$
• 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 14:40
• Ｉmean to solve the equation $\left(\frac{a}{x}\right)^{x+1}=b$. Mar 21, 2020 at 15:32
• 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 0:43