Unknown both as a exponent and as a term in an equation Let's say I have an equation $e^{x-1}(x+1)=2$. According to Solving an equation when the unknown is both a term and exponent it's impossible to solve this using elemetary functions. If so, then how do you solve it? Could you give me some keywords for further research on solving this type of equations? I'd be even more grateful if you solved the equation above (if it's not too hard).

Seems to be a case for the Lambert W-function:
\begin{align}
e^{x-1}(x+1) &=2 \iff \\
e^{x+1}(x+1) &=2e^2 \iff \\
f(x+1) &= 2e^2 \Rightarrow \\
x + 1 &= f^{-1}(2e^2) \Rightarrow \\
x &= f^{-1}(2e^2) - 1
\end{align}
where $f(x) = x e^x$ and $f^{-1} = W$, which is that Lambert W-function.
We need the branch with positive real numbers, which is called $W_0$:
$$
x = W_0(2e^2) + 1
$$
Using $W_0(x e^x) = W_0(f(x)) = x$ we see $W_0(2e^2) = 2$ and
$$
x = 2 - 1 = 1
$$
Note: This particular problem only needed the existence of an inverse to $f(x) = x e^x$.
We could have written
$$
x = f^{-1}(2e^2) - 1 = f^{-1}(f(2)) - 1 = 2 - 1 = 1
$$
without knowing about $W_0$. The existence of an inverse could be argued by the monotonicity of $f$.
