What is the function $f$ such that $(x + f(x))^\alpha = x^\alpha + a$?

Question

What is the function $f$ such that $(x + f(x))^\alpha = x^\alpha + a$, with $\alpha,a > 0$.

Answer 1

Applying the logarithm to both sides of the equation we have: $$ \ln{(x+f(x))}=\frac{\ln{(x^{\alpha}+a})}{\alpha} \Rightarrow $$ $$ f(x) = (x^{\alpha}+a)^{1/\alpha} - x $$

Added

The domain of $f$ depends on the value of $\alpha$. e.g. $$ \alpha = n\in \mathbb{N} \Rightarrow \text{Dom}(f) = (-\infty,\infty) $$ We are assuming an unprescribed "precalculus" domain that "works", i.e. the output: $f(x)$ will be a real number given any $x \in \text{Dom}(f)$ .

Or like Haskell said in the comments, a common blanketed domain that "works" for all $\alpha > 0$ would be: $\text{Dom}(f) = [0,\infty)$. I like this, and think this is more satisfactory.