Solving $x=\frac{2^{1+y}}{\left(y+1\right)\left(y+2\right)}$ for $y$ Do you know how to solve the following equation to make $y$ the subject:
$$x=\frac{2^{1+y}}{\left(y+1\right)\left(y+2\right)}$$
My attempts:
I multiplied both sides by $\ln(2)(y+1)^2$ to get
$$x\ln\left(2\right)\left(y+1\right)^{2}\left(y+2\right)=\ln\left(2\right)\left(y+1\right)e^{\ln\left(2\right)\left(y+1\right)}$$
The reasoning behind this was to get an expression of the form $xe^x$ to use the Lambert W function. However I was unable to make it work. I also tried considering the more general function
$$x=\frac{z^{1+y}}{\left(y+1\right)\left(y+2\right)}$$
Differentiating both sides with respect to z cancels the $(y+1)$ term in the denominator which I thought could make it easier to work with. However again I was unable to find a solution.
Any help is appreciated. Thanks
 A: $$x=\frac{2^{1+y}}{(y+1)(y+2)}$$
$$x=\frac{e^{\ln(2)(1+y)}}{(y+1)(y+2)}$$
$y\to t-1$:
$$\frac{1}{t(t+1)}e^{\ln(2)t}=x$$
We cannot read from the equation how to invert it by elementary functions:
How can we show that $A(z,e^z)$ and $A(\ln (z),z)$ have no elementary inverse?
We see, as already stated in the other answers, this equation isn't solvable by elementary functions and Lambert W. But it is solvable by elementary functions together with Generalized Lambert W.
$$t=\frac{1}{\ln(2)}W\left(^{\ \ \ \ \ \ -}_{0,-\ln(2)};\frac{x}{\ln^2(2)}\right)$$
$$y=\frac{1}{\ln(2)}W\left(^{\ \ \ \ \ \ -}_{0,-\ln(2)};\frac{x}{\ln^2(2)}\right)-1$$
[Mezö 2017] Mezö, I.: On the structure of the solution set of a generalized Euler-Lambert equation. J. Math. Anal. Appl. 455 (2017) (1) 538-553
[Mezö/Baricz 2017] Mezö, I.; Baricz, A.: On the generalization of the Lambert W function. Transact. Amer. Math. Soc. 369 (2017) (11) 7917–7934 (On the generalization of the Lambert W function with applications in theoretical physics. 2015)
[Castle 2018] Castle, P.: Taylor series for generalized Lambert W functions. 2018
