Question regarding complex numbers and real numbers? I have two questions...
If we take $(-1/3)^{(-1/3)}$ it would equal $-1.44224957$ since...
$$(-1/3)^{-1/3}$$
$$\frac{1}{(-1/3)^{(1/3)}}$$
$$\frac{1}{-0.6933612744}$$
$$-1.44224957\ldots$$
Yet when I place my calculator as $$e^{\left(-\frac{1}{3}\right)\ln\left(-\frac{1}{3}\right)}$$
I would get the complex number of $$.721124785-1.124902477i$$
So does $.721124785-1.124902477i=-1.44224957$?
As I continued with this I found that for negative $x$-values $x^x=(-x)^x$ for $x$ values with $-\frac{\text{even numerators}}{\text{odd denominators}}$ and $-(-x)^{(x)}$ for x values for $-\frac{\text{odd numerator}}{\text{odd denominators}}$ so....
$$(-x)^x=e^{x\ln(-x)}$$
$$-(-x)^x=-e^{x\ln(-x)}$$
For the both of them...
$$\ln{\left((-x)^{x}\right)}=x\ln(-x)$$
Then dividing by $x$
$$\ln(-x)=\ln(-x)$$
So does $\ln(x)=\ln(-x)$ when the $x$-value is $-\frac{\text{even/odd number}}{\text{odd number}}$?
 A: It will never be true that $\ln(-x) = \ln(x)$ because the logarithm of a negative must always be complex.  It's $\require{cancel} \cancel{\text{possible}}$ that there are complex solutions of $\text{Log}(x)$ that equal a complex solution of $\text{Log}(-x)$.
I think it's easiest to solve this in the context of general complex numbers, i.e. $re^{\theta i}$.  Now we have the following:
$$
\text{Log}\left(re^{\theta i}\right) = y \rightarrow e^y = re^{\theta i + 2\pi ni}
$$
We can safely rewrite $r = e^{\ln(r)}$ since $r$ is a positive real number which gives:
$$
e^y = e^{\ln(r) + \theta i + 2\pi ni}
$$
Finally we see that $\text{Log}(re^{\theta i}) = \ln(r) + (\theta + 2\pi n)i$.  When $x$ is a real positive value we get $r = x$ and $\theta = 0$:
$$
\text{Log}(x) = \ln(x) + 2\pi ni
$$
When $x$ is negative we get $r = x$ and $\theta = \pi$:
$$
\text{Log}(-x) = \ln(x) + (1 + 2n)\pi i
$$
So you can see that there are no values of the general Logarithm where it gives the same value for a positive and negative real value (since the negative value always requires odd integers and the positive value always requires even integers).
Note
You might wonder about that addition by $\ln(r)$ (remember $r$ is a positive real number).  Can't we write $r = e^{\ln(r)}$ with the more general $r = e^{\text{Log}(r)}$?  Well, of course you can, but then we simply get:
$$
\text{Log}\left(re^{\theta i}\right) = \ln(r) + 2\pi mi + \theta i +  2\pi n i
$$
But this ends up being $\text{Log}\left(re^{\theta i}\right) = \ln(r) + \theta i + 2\pi(m + n)i$...an addition of two integers is still just an integer though.
