Inverse function of $x+n^x$ I need to find the inverse function of $f(x)=x+n^x$, where $n$ is a variable in the range $(0, 1]$. I understand that the result would not  be a function because of the vertical-line test, so it would need  to be broken up.
By swapping $x$ for $y$ we get $x=y+n^y$, but I have no idea where to go from here to isolate $y$ on one side of the equation.
 A: Starting with $x=y+n^y$, and manipulating the equation in some non-obvious ways:
\begin{align*}
x&=y+n^y\\[5pt]
\implies x-y &=n^y\\[5pt]
\implies (x-y)\ln(n)n^{x} &= n^{y}\ln(n)n^{x}\\[5pt]
\implies(x-y)\ln(n)n^{x-y} &= n^{x}\ln(n)\\[5pt]
\implies \ln\left(n^{x-y}\right)n^{x-y} &= n^{x}\ln(n)\\[5pt]
\implies \ln\left(n^{x-y}\right) \exp\left(\ln\left(n^{x-y}\right)\right) &= n^{x}\ln(n)\\[5pt]
\implies W(n^{x}\ln(n)) &= \ln\left(n^{x-y}\right)\\[5pt]
\implies W(n^{x}\ln(n)) &=(x-y)\ln(n)\\[5pt]
\implies W(n^{x}\ln(n)) &=x\ln(n) - y\ln(n)\\[5pt]
\end{align*}
and so by solving the last equation for $y$ we get $$\boxed{y = x - \frac{W(n^{x}\ln(n))}{\ln(n)}}$$ where $W$ is the Lambert W function.
A: Equations of the form $a^x+cx=b$ are in general solvable using the Lambert W function, the inverse of $xe^x$. In your case, if $y=x+n^x$ then
$$x=y-\frac{W(n^y\ln n)}{\ln n}$$
Choosing different branches of $W$ gives all possible values of $x$ given $y$.
A: $$y(x)=x+A^{\,x}\qquad A=\text{constant}$$
$$y-x=A^x=A^{x-y}A^y$$
$$(y-x)A^{y-x}=A^y$$
$$(y-x)e^{(y-x)\ln(A)}=e^{y\ln(A)}$$
$$(y-x)\ln(A)e^{(y-x)\ln(A)}=\ln(A)e^{y\ln(A)}$$
Let $\quad X=(y-x)\ln(A)\quad$ and $\quad Y=\ln(A)e^{y\ln(A)}$
$$Xe^X=Y$$
The inverse function $X(Y)$ cannot be written with a finite number of elementary functions. A special function is required, namely the Lambert W function.
https://en.wikipedia.org/wiki/Lambert_W_function
$$X=W(Y)$$
Note that this function is multivalued, namely $W_0(Y)$ and $W_{-1}(Y)$ in reals and more determinations in complex.
$$(y-x)\ln(A)=W\left(\ln(A)e^{y\ln(A)} \right)$$
Thus the formal answer to your question is :
$$x=y-\frac{1}{\ln(A)}W\left(\ln(A)e^{y\ln(A)} \right)$$
$\boxed{\text{I see that Parcly Taxel and DMcMor gave the answer while I was still typing.}  }$
A: If you consider the function $f(x)=x+n^x$ from $\mathbb{R} \to \mathbb{R}$ notice that since $f(x)=n^x+x$ is not surgetive for some  $n\in (0,1]$, the function don´t should be inverse function for all $n$.
Is clear that for $n=1$ the function have inverse, but for arbitrary $n$ you can´t afirm these.
