In school I have been taught that $\log_a\frac{r}{s} = \log_ar - \log_a{s}$. Furthermore, it has been said that this rule can be inverted (applied from the right to the left). Then, I came across this:
Simplify $\lg(x-y) - \lg(y-x)$.
Note that $\lg(x)$ means $\log_{10}(x)$. Let's consider $x, y \in \mathbb R^+$.
The "simplification" yields $\lg\frac{x-y}{y-x} = \lg(-\frac{y-x}{y-x}) = \lg(-1) \space\space\space ( = \frac{\ln(-1)}{\ln(10)} = \frac{\pi i}{\ln(10)})$.
However, if you look at the task, you would get different results depending on whether $x > y$ or $y > x$. (try it with a calculator or simply google an example...) So, is this rule wrong or do I have to make these "case differentiations" every time I am using this rule?
This is one of those cases where there are two seemingly different expressions that express the same value. $\mathbb e ^ {\mathbb i \pi} = -1$ and $\mathbb e ^ {-\mathbb i \pi} = -1$. Both are valid solutions to $\lg\frac{x-y}{y-x}$ and which one you arrive at depends on how you break down your expression.