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

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On the surface, $f(x) = (x^{\alpha} + a)^{1/\alpha} - x$, but we really know nothing about the domain of $f$. – Ron Gordon Jan 30 '13 at 0:03
As $ \alpha,a > 0 $, we can assume that $ \text{Dom}(f) = [0,\infty) $. – Haskell Curry Jan 30 '13 at 0:05
@Sh3ljohn: It is impossible to deduce the existence of $ f $ without knowing what it looks like in the first place. Hence, we are assuming that there exists a function $ f: [0,\infty) \to \mathbb{R} $ such that $ x + f(x) \in [0,\infty) $ and $ (x + f(x))^{1/\alpha} = x^{\alpha} + a $ for all $ x \in [0,\infty) $. Such an $ f $ must be of the form given by rlgordonma and Rustyn. We then verify that $ f $ is indeed a valid solution. – Haskell Curry Jan 30 '13 at 0:25
@Rustyn: Yes. I was just assuming a common domain that works for all $ \alpha > 0 $. :) – Haskell Curry Jan 30 '13 at 0:26
up vote 5 down vote accepted

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


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.

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Thanks and sorry for my comment up there, I didn't understand it at first. – Sh3ljohn Jan 30 '13 at 0:22

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