logarithm of the product of non positives numbers

Question

According to wikipedia: $${\displaystyle \ln(xy)=\ln x+\ln y\quad {\text{for }}\;x>0\;{\text{and }}\;y>0}$$

Does this formula also hold if say $x<0$ and $y>0$ ?

Answer 1

In the context of real numbers, no, since a non-positive real numbers has no logarithm.

Answer 2

Let's consider the definition of a logarithm. Note that $b$ is a positive real number.

$$\log_b{(x)}=y \Rightarrow b^{y}=x$$

If we are considering real numbers, then no combination of $b$ raised to $y>0$ will yield an $x<0$.

Hence, in the context of real numbers, the condition $\ln(xy)=\ln(x)+\ln(y)$ holds for $x>0$ and $y>0$.

Answer 3

Considering you are asking about the logarithm of negative numbers, I assume you are familiar with the complex logarithm. Since $e^{i\pi} = -1$, it makes sense to assign $\ln(-1) = i\pi$ and indeed, one can make this rigorously, by taking $\ln$ to be the inverse of $$\exp : \left\{(x+iy) \in \mathbb{C} : x \in \mathbb{R}, y \in \left(-\frac{\pi}{2},\frac{3\pi}{2}\right)\right\} \to \mathbb{C} \setminus \{-iy : y \in \mathbb{R}_+\}.$$ This gives the correct value for positive real numbers and also assigns $\ln(-1) = i\pi$. But even then, it does not hold $\ln(ab) = \ln(a) + \ln(b)$ anymore, consider e.g. $a = b = -1$.