Inverse function of $x+n^x$

Question

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.

Answer 1

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.

Answer 2

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 $y$.

Answer 3

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.

Answer 4

$$y(x)=x+A^{\,x}\qquad A=\text{constant}$$ $$y-x=A^x=A^{x-y}A^y$$ $$(y-x)A^{y-x}=A^y$$ $$(y-x)e^{(y-x)\ln(A)}=e^{y\ln(A)}$$ $$(y-x)\ln(A)e^{(y-x)\ln(A)}=\ln(A)e^{y\ln(A)}$$ 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 : $$x=y-\frac{1}{\ln(A)}W\left(\ln(A)e^{y\ln(A)} \right)$$

$\boxed{\text{I see that Parcly Taxel and DMcMor gave the answer while I was still typing.} }$