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 Oct 28 comment How to make $\log x^a = a\log x$ work using multivalued complex approach So if $f(1) = \{0, 1\}$, we established that $2f(1) = \{0, 1, 2\}$. What is $\frac{1}{2} f(1)$ equal to? Oct 5 comment How to make $\log x^a = a\log x$ work using multivalued complex approach I have a Python code that tests those identities here: theoretical-physics.net/dev/math/… Oct 4 comment How to make $\log x^a = a\log x$ work using multivalued complex approach For example, what if $a={1\over2}$? What are all the possible values for ${1\over2}\log z$? Oct 3 comment How to make $\log x^a = a\log x$ work using multivalued complex approach I think you are right! Do you have some references for how to do arithmetic with multivalued functions? E.g. what if "a" is not an integer (I know that's not the particular example I posted)? Oct 3 revised How to make $\log x^a = a\log x$ work using multivalued complex approach This corrects a mistake in derivation, it is not true for all 'z' that Log(e^z) = z Oct 3 comment How to make $\log x^a = a\log x$ work using multivalued complex approach Just in case you can't see my edit, the final equation that you get, using your notation, is: $\log(k,x^a)=a\log(l, x)+2\pi i \left(k - la + \left\lfloor \pi-\Im (a\Log(x)) \over 2\pi \right\rfloor\right) ,k,l\in\mathbb{Z}$ Oct 3 comment How to make $\log x^a = a\log x$ work using multivalued complex approach You need to be careful, it is not true for all $z$ that $\Log(e^z) = z$, but rather it is true for all $z$ that $\Log(e^z) = z + 2\pi i \left\lfloor \pi-\Im z \over 2\pi \right\rfloor$. I have proposed edits to your answer that fixes this problem (I don't know if you can see them or not). Either way, I gave you the bounty for your work, but it seems that the answer is that it is only possible if you use the floor function. @dafinguzman, yes, your winding number is precisely the floor function in all my equations. Oct 3 suggested approved edit on How to make $\log x^a = a\log x$ work using multivalued complex approach Oct 3 awarded Benefactor Oct 3 comment How to make $\log x^a = a\log x$ work using multivalued complex approach I think you made a mistake in equation (1), it should rather be $a\log(x) = a(\Log(x) + 2\pi i n) = a\Log(x) + 2 a\pi i n$, so then the corrected (1) and your (2) are not equal. Oct 2 comment How to make $\log x^a = a\log x$ work using multivalued complex approach Thanks for the answer. Yes, if you have to choose a specific branch, e.g. the principal branch with $\Log(z)$ as in your quote from the book, then the only way to have a formula valid for all $x$ and $a$ is to use the very first equation in my question: $\Log x^a = a\Log x + 2\pi i \left\lfloor \pi-\Im (a\Log x) \over 2\pi \right\rfloor$, which always holds. However, I thought that somehow you can make the "more natural" formula $\log x^a = a\log x$ work with multivalued $\log(z)$, perhaps by some convention how to evaluate it. Your answer seems to suggest that there is no such thing. Oct 1 comment How to make $\log x^a = a\log x$ work using multivalued complex approach I am interested in how to prove the relation $\log x^a = a \log x$ where $\log z$ is a multi-value analytic function. If you don't like $x=-1$, then use $x=1$. The result doesn't change, there are some values on LHS that are not included in the RHS. Sep 30 comment How to make $\log x^a = a\log x$ work using multivalued complex approach The problem is that there are are members of LHS (e.g. $n = 1$) that are not members of RHS (no matter what you choose for $m$). If that's the case, how can LHS be equal to RHS? Or to be specific, in what exact sense is it equal? Sep 28 comment Principal Value of complex log You made a mistake there, $\text{Log}^i\,(1-\sqrt3\,i) \neq e^{i\,\text{Log}\,(1-\sqrt3\,i)}$, but rather $\text{Log}^i\,(1-\sqrt3\,i) = e^{i\,\text{Log} \text{Log}\,(1-\sqrt3\,i)}$. Sep 27 awarded Promoter Sep 25 asked How to make $\log x^a = a\log x$ work using multivalued complex approach Dec 16 awarded Caucus Jul 2 awarded Curious Nov 20 accepted Derivative of big O symbol Nov 18 revised Derivative of big O symbol Clarify the x=0 point once more, just to be absolutely clear we are not dealing with x=oo