just a quick question with the hopes someone may be able to help clarify where my confusion is.
I've been asked to find which equality or conclusion is wrong. we have
$$2Log(2) = Log(2^2) = Log((-2)^2)=2Log(-2)$$
now, the important part of this question i believe is to consider which branch we're on. and so i include the following two definitions from my source material
Def: We define the Arg(z), the principal value of the arguement as $$Arg(z) = \theta \longleftrightarrow |z|(cos \theta + i sin \theta), ~\theta \in [0,2\pi)$$
and
Def: We define the principal branch Log of the log function to be $$Log(z) = log|z| + i arg(z),~ 0 \leq arg(z) < 2\pi$$
Now, $$2Log(-2) = 2\pi i + Log(4)$$ i believe But we have to note the use of the Big L, i know we're on the principal branch, from my memory then, for the fact the range is $[0,2\pi)$ then we're on the branch $ z \in [0,\infty)$ specifically. in which case, does that make $2 Log(-1)$ undefined? in which case can i negate that entirely leading to a true equality between the two?
and so i have a better understanding if we were on the branch $z \in (-\infty,0]$ we'd be using $-\pi < arg \leq \pi$ in which case, we wouldnt be using the principal branch (as defined above) of Log(4) since that would be undefined here.
Thanks for the help, Sincerely.
There's no need to be bothered much about what's happening -- it's not farfetched. The equality $$\log a^b=b\log a$$ fails to remain true whenever $a$ is negative since then LHS is single-valued, finitely valued or infinitely valued (depending on whether $b$ is respectively an integer, a nonintegral rational number, or an irrational number) whereas RHS is always multi-valued in an infinite manner. Even at that it's not guaranteed that there are some values on either side that will make the equality true. Consider the case of wanting $$\log(-1)^2=2\log(-1)$$ to be true, for example. Since RHS is $2(iπ+2πik)$ for an arbitrary integer $k,$ we see that there is no $k$ so that $$0=2πi(1+2k),$$ so that in this case the equality is false through and through.
Thus the problem is in the last step where you write $$\log(-2)^2=2\log (-2).$$ This is plain false since in this case LHS has a single value whereas RHS has infinitely many values, for one. This is only true if there is a choice of one of the values on RHS that becomes equal to $\log 4,$ so that we may restrict the equation to such a case, and in that case only is it true.