Is $\log(-x)$, where $x \in (0, \infty)$ undefined? From solving quadratics I was first told that the discriminant has to be non-negative, since $\sqrt{-a}$, where $a \in (0, \infty)$ is undefined. But this was before learning about Imaginary numbers and the complex plane.
For example, what would $\log_3(-9)$ be? Is there a number $a$ such that $3^a=-9$?
So can you evaluate logarithms of negative numbers?
Yes, it is possible to evaluate logarithm of a negative number in the complex plane. Moreover it is possible evaluate logarithm of any non-zero complex number $z=x+iy\ne0$: $$ \log z=\ln|z|+i\arg(z). $$ where the real numbers $|z|=\sqrt{x^2+y^2}$ and $\arg z$ are, respectively, the absolute value and argument of $z$. The argument is essentially the angle in the complex plane between $z$ and positive direction of the real axis. There is however a complication. Different from the real logarithm the complex one is multivalued function, so that any multiple of $2\pi i$ can be added to its value. One of possible solution of the problem is to bound the imaginary part (for example from $-\pi$ to $\pi$).
Equipped with this knowledge and the fact that $\log_a z=\frac{\ln z}{\ln a}$: $$ \log_3(-9)=\frac{\ln9+i\pi}{\ln3}=2+\frac{\pi i}{\ln3}. $$