Why aren't logarithms defined for negative $x$? Given a logarithm is true, if and only if, $y = \log_b{x}$ and $b^y = x$ (and $x$ and $b$ are positive, and $b$ is not equal to $1$)&#x5B;1&#x5D;, are true, why aren't logarithms defined for negative numbers?

Why can't $b$ be negative? Take $(-2)^3 = -8$ for example. Turning that into a logarithm, we get $3 = \log_{(-2)}{(-8)}$ which is an invalid equation as $b$ and $x$ are not positive! Why is this?
For the real, continuous exponentiation operator -- the used in the definition of the real, continuous logarithm -- $(-2)^3$ is undefined, because it has a negative base. 

The motivation stems from continuity: If $(-2)^3$ is defined, then $(-2)^{\pi}$ and $(-2)^{22/7}$ should both be defined as well, and be "close" in value to $(-2)^3$, because $\pi$ and $22/7$ are "close" to 3. But the same line of reasoning says that $(-2)^{22/7}$ should be  $8.8327...$, which isn't very close to $-8$ at all, and what could $(-2)^{\pi}$ possibly mean?!?!

You can define a "discrete logarithm" operator that corresponds to the discrete exponentiation operator (i.e. the one that defines $(-2)^3$ to be repeated multiplication, and thus $8$), but this has less general utility -- my expectation is that one would be much better off learning just the continuous logarithm and dealing with the signs "manually" in those odd cases where you want to solve questions involving discrete exponentiation with negative bases. (and eventually learn the complex logarithm)
In $\log_{−2}(8) = 1/2$, the way to solve it normally is to write: $(-2)^{1/2} = 8$.
But in this case, $(-2)^{1/2}$ is undefined. Hence, negative bases for log cannot be used.