Why does $\log\frac{x}{x-2}$ not equal $\log(x) - \log(x-2)$

Question

Based on log law: $$\log \left( \frac{a}{b} \right) = \log(a)-\log(b) \ ,$$ why does $$\log \left( \frac{x}{x-2} \right) \neq \log(x) - \log(x-2) \ ?$$

I have found that the graph of $\log \left( \frac{x}{x-2} \right)$ is defined for negative $x$ values while $\log(x) - \log(x-2)$ does not. Why is this? Sorry if this is a simple question.

Answer 1

Try $x=-1$.

You'll get something wrong.

For $x>2$ we have $$\ln\frac{x}{x-2}=\ln{x}-\ln(x-2)$$ and for $x<0$ we have $$\ln\frac{x}{x-2}=\ln(-x)-\ln(2-x)$$

For $a>0$ and $b>0$ we obtain: $$\ln\frac{a}{b}=\ln{a}-\ln{b}.$$

Answer 2

For $\ln(\frac{x}{x-2})$ to defined $$\frac{x}{x-2}\gt 0 $$ $$x\in(-\infty,0)\cup(2,\infty)$$ Only in this interval $$\ln(\frac{x}{x-2})=\ln(\vert x\vert)-\ln(\vert x-2 \vert),\ \ x\lt 0 ,\ x\gt 2 $$

Answer 3

Easy. Because your log law is valid only for positive arguments. That is $$\log(a/b)=\log a -\log b$$ is true only for positive values of $a$ and $b,$ for otherwise, $\text{RHS}$ is no longer a definite value, even if $\text{LHS}$ is. In any case one then sees that the equation is false. This happens because a quotient of two negatives is positive. So it's all a consequence of the $\text{minus}×\text{minus}=\text{plus}$ stuff.