Is $ \log_a x^n = n \log_a x? $ We have:
$$\log_a{x^n}=\log_a{\left( x\cdot x \cdot x \cdot ...\cdot x\right)}=\log_a(x)+\log_a(x)+...+\log_a(x)=n\log_a(x)$$
But,
$$ \ln\left(-1\right)^2=\ln{\left(-1\right)^2}\\\ln(1)=2\ln(-1)\\ 0=2i\pi$$
or
$$ \log_a(-1)^3=\log_a(-1)^3 \\ \log_a(-1)=3\log_a(-1) \\1=3 $$
So, if the "identity" first written is just for $x\in\mathbb{R}| x>0$ tell me why, or where I'm I failing at!
 A: If you'd like to use the identity
$$
\ln x^n=n \ln x,
$$
you should consider the definition of logarithm as a multivalued function:
$$
\textrm{Ln}\, z=\ln |z|+i \mathrm{Arg}\, z=\ln |z|+i(\theta+2\pi k),
$$
i.e., there are infinitely many values. Using this definition your examples are valid:
$$
\textrm{Ln}\,(-1)^2=\textrm{Ln}\, (-1)^2\\
\textrm{Ln}\,1=2 \textrm{Ln}\, (-1)\\
i 2\pi k=2i(\pi+2\pi m),
$$
where the last equality is understood modulo $2\pi k i$.
A: The identities 
$$\log_a{x^n}=n\log_a(x)$$
and
$$\log_a(xy)=\log_a(x)+\log_a(y) \,.$$
are proven for positive $x,y$. Their proof uses (hidden) the fact that the exponential function $a^x : \mathbb R \to (0, \infty)$ is a bijection....This is not true anymore when you extend it to complex numbers.
A: The exponentiation function is not one-to-one.  For any $a$ and $x$, there are many complex numbers $y$ such that $$a^y = x$$ and therefore there are many numbers $y$ that of which it can be said that $$y=log_a x.$$ For positive real $x$ we can define $y$ to be the unique real number for which this holds; for other $x$, we cannot.
This means that you cannot conclude from $$\log_a x = \log_a x'$$ that $$x = x'$$
except under certain limited circumstances, just as you cannot conclude from $$x^2 = y^2$$ to $$x = y$$ except under certain limited circumstances.
