How to make $\log x^a = a\log x$ work using multivalued complex approach The following identity holds for all $a$ and $x$ using the principal branch:
$$
    \log x^a = a\log x
        + 2\pi i \left\lfloor \pi-\Im (a\log x) \over 2\pi \right\rfloor
$$
e.g. for $a=2$, $x=-1$:
LHS = $\log (-1)^2 = \log 1 = 0$
RHS =
$2\log(-1)+ 2\pi i \left\lfloor \pi-\Im (2\log (-1)) \over 2\pi \right\rfloor$
=$2\pi i+ 2\pi i \left\lfloor \pi-\Im (2\pi i) \over 2\pi \right\rfloor = 2\pi i - 2\pi i = 0$
Everything is single valued and there is no problem.
How can I perform the same calculation using multivalued logarithms? In other words, I want to use the (multivalued) identity:
$$\log x^a = a\log x$$
for $a=2$, $x=-1$. I pick the same branch for both LHS and RHS, let's pick the principal branch (so that we can reuse the values calculated above), and then I add the $2\pi i n$ term for each logarithm and get:
LHS = $\log (-1)^2 + 2\pi i n = 2\pi i n$
RHS = $2\log(-1) + 2\pi i \left\lfloor \pi-\Im (2\log (-1)) \over 2\pi \right\rfloor + 2\cdot 2\pi i m = 4\pi i m$
And we can see that LHS is not equal to RHS, otherwise $m$ would have to be half-integer.
Where did I make the mistake? How can I make LHS equal to RHS using the multivalued approach?
 A: $\DeclareMathOperator{\Log}{Log}$
Here is what Churchill, Brown, Verhey say about this particular case (Complex Variables and Applications, p. 66, Third Edition) and I quote:
"The statement $\log(z^n)=n\log(z)$ may or may not be true for specific values of $z$ and $n$ when the multiple-valued complex logarithmic function is replaced by a single branch of it. If we use the notation $\Log$ for the principal branch of the complex logarithmic function, note for example, that $\Log[(1+i)^2]=2\Log(1+i)$ while $\Log[(-1+i)^2]\neq 2\Log(-1+i)$.",
while you write (in the comments): "I am interested in how to prove the relation $\log(x^a)=a\log(x)$ where $\log(z)$ is a multi-value analytic function."
With your case, $\Log(x^a)=a\Log(x)$ holds for the specific values $x=-1$ and $a=2$.
Consequently the values of $x$ and $a$ do matter, so you are trying to prove something that may not be true in general.
Addendum #1 (after your comment):
There is such a thing, but you have to be really careful to make a clear distinction between the two symbols: $\log(z)$ and $\Log(z)$. Because $\log$ is multivalued and indexed by $k\in\mathbb{Z}$, it is more convenient to use the following notation for this map:
$$\log(k,z)=\Log(z)+2k\pi i,\,\,k\in\mathbb{Z}\Rightarrow$$
Using the fact that:
$$\Log(e^z) = z + 2\pi i \left\lfloor \pi-\Im z \over 2\pi \right\rfloor$$
we get:
$$\log(k,e^z)=z+2\pi i\left(k + 
\left\lfloor \pi-\Im z \over 2\pi \right\rfloor
\right),\,\,k\in\mathbb{Z}$$
If you now set $z=x^a$ to the the above, using the principal branch of the logarithm to define it as: $x^a=e^{\Log(x^a)}$, you get the (multi-valued) set constructor you are looking for:
$$\log(k,x^a)=\log(k,e^{\Log(x^a)})=\Log(x^a)+2\pi i
\left(k + \left\lfloor \pi-\Im \Log(x^a) \over 2\pi \right\rfloor\right)
,k\in\mathbb{Z}$$
You cannot reduce the $\Log(x^a)$ above to $a\Log(x)$ (no matter what the rest of the expression is), because that would be using the identity $\Log(x^a)=a\Log(x)$, which as Churchill shows, might not hold for the principal branch and the particular values of $x$ and $a$ you are considering.
However, using the definition of a power $x^a = e^{a\Log x}$, we obtain:
$$\log(k,x^a)=\log(k,e^{a\Log(x)})=a\Log(x)+2\pi i
\left(k + \left\lfloor \pi-\Im (a\Log(x)) \over 2\pi \right\rfloor\right)
,k\in\mathbb{Z}$$
and using $a\Log x = a\left(\log(l, x) - 2\pi l i\right)$ we get:
$$\log(k,x^a)=a\log(l, x)+2\pi i
\left(k - la + \left\lfloor \pi-\Im (a\Log(x)) \over 2\pi \right\rfloor\right)
,k,l\in\mathbb{Z}$$
A: $log(x) = Log(x) + i2\pi n = Log(|x|) + iArg(x) + i2\pi n$ where $Log$ and $Arg$ are principal values.
$log(x) + log(x) = 2Log(|x|) + i2Arg(x) + i2\pi n + i2\pi m$
$n$ and $m$ are independent and can be merged into one variable $k$.
$alog(x) = aLog(|x|) + iaArg(x) + i2\pi k$
$$alog(x) = aLog(x) + i2\pi k\ \ \ (1)$$
$e^{alog(x)} = e^{aLog(x) + i2\pi k} = x^ae^{i2\pi k} = x^a$ (sanity check)
$$log(x^a) = Log(x^a) + i2\pi k = aLog(x) + i2\pi k\ \ \ (2)$$
Equations (1) and (2) are equal.
$$log(x^a) = alog(x) = aLog(x) + i2\pi k\ \ \ (3)$$
Also introduce z for clarity.
$alog(z) = aLog(x) + i2\pi k$
$log(z) = Log(x) + i\frac{2\pi k}{a}$
Take the exponential.
$z = xe^{i\frac{2\pi k}{a}}$
$e^{i\frac{2\pi k}{a}}$ are roots of unity.
How to do the calculation:
$log((-1)^2) = Log(1) + i2\pi k = i2\pi k$
$2log(-1) = 2Log(-1) + i2\pi n = 2(i\pi) + i2\pi n = i2\pi (n+1)$
$k = n+1$
Are multi-valued functions a rigorous concept or simply a conversational shorthand?
A: $$log(z^\frac{p}{q}) = \frac{p}{q}log(z) = \frac{p}{q}(Log(z) + i2\pi k) + i2\pi n$$
 The justification I recall is that the exponential of the LHS equals that of the RHS. The derivation went along the lines of:
Replace $z$ with $ze^{i2\pi k}$. Let $y=(ze^{i2\pi k})^\frac{p}{q}$. 
$$log(y) = Log(y) + i2\pi n = Log((ze^{i2\pi k})^\frac{p}{q}) + i2\pi n = \frac{p}{q}Log(ze^{i2\pi k}) + i2\pi n$$
$$log(y) = \frac{p}{q}(Log(z) + i2\pi k) + i2\pi n$$
