If you posit $\log(-x)=\gamma \log(x)$ for all $x$, and if you want to allow the usual operations like division, you are going to be forced to conclude that $\gamma=\log(-x)/\log(x)$ for all $x$, and in particular that
$$\frac{\log(-2)}{\log(2)}=\frac{\log(-3)}{\log(3)}$$
But the various logs in this equation already have definitions, and according to those definitions, the equation in question is not true (for any of the various choices of $\log(-2)$ and $\log(-3)$.
Therefore, your $\gamma$ can exist only if you either ban division or change the definition of the log. Likewise for your other proposal $x^{\gamma k}=-x^k$.
This is why you can't just go adjoining new constants willy-nilly and declaring them to have whatever properties you want. In the case of $i$, the miracle is that you can define it in a way that does not require you to revise the existing rules of arithmetic. Such miracles are rare.
The concept that you are missing is the Complex plane. Not surprisingly, given that I often seen it introduced a full year after imaginary numbers.
On the complex plane, taking a square root becomes almost entirely geometric. The non geometric part is, you have to take the square root of the distance from your number to 0. For instance $15+20i$ is at a distance of $25$ from 0 (per Pythagorean Theorem) so the distance of the square root will be 5.
To find the angle of the square root—just note the angle of the original number (measuring counter-clockwise from the real axis—a.k.a. the x axis) and cut it in half.
You may notice that there are two ways to cut the angle in half—clockwise, and counter-clockwise. That is correct; every number except 0 has two square roots.
You asked about negative numbers. Let's take $-i$ as an example. Note that the position of $-i$ on the complex plane is one unit "south" of 0, the same place that $(0,-1)$ is on a rectangular coordinate plane. Since the distance from 0 is 1, and the positive square root of 1 is 1, the distance from 0 for our answer will also be 1, making that part easy.
Note that the angle of that point, measured counter-clockwise from the x axis, can be stated as either $270$ or $-90$. Use both. Divide each by two. That gives the two answers.
Now solving with the pythagorean theorem (or using trig, but in this case the angles are just 45 degrees in the 2nd and 4th quadrants, so we don't need trig), we get the two square roots of $-i$, which are $\frac{1-i}{\sqrt2}$ and $\frac{-1+i}{\sqrt2}$.
Try squaring each of those with pencil and paper to convince yourself they are indeed correct.
Best Answer
Every non-zero complex number $z$ has exactly two complex square roots - this is a consequence of the field of complex numbers being algebraically closed (Wikipedia link). If $$z=re^{i\theta}=r\cos(\theta)+ri\sin(\theta)$$ then the square roots of $z$ are $$\begin{align*} \sqrt{r}e^{i\theta/2}&=\sqrt{r}\cos(\theta/2)+\sqrt{r}\,i\sin(\theta/2)\\ -\sqrt{r}e^{i\theta/2}&=-\sqrt{r}\cos(\theta/2)-\sqrt{r}\,i\sin(\theta/2) \end{align*}$$ In short, there are no complex numbers whose square roots are not already present in the complex numbers themselves, so there is nothing more to "add in".