Compare between $\ln(2)$, $\ln(-2)$

Question

$x^2 =(-x)^2, \;\forall x \in \mathbb{R}^+$

$$\begin{align*} \therefore \ln(x^2) &=\ln(-x)^2\\ 2\ln(x)&=2\ln(-x)\\ \ln(x)&=\ln(-x) \end{align*}$$

If the statement above is correct, then compare between: $\ln(2)$ and $\ln(-2)$?

My try is, I think the statement is wrong, because $x$ must be positive? any help?

By the way, named math operators should appear upright, and the common ones have their own code for this purpose (e.g. \ln, \sin - see entry 11 in our MathJax guide). — Zev Chonoles, Jul 08 '15 at 03:57
$\ln(-x)^2$ is not clear in precedence. It would be better to write $\ln((-x)^2)$ to distinguish from $(\ln(-x))^2$. This does not change your argument. — Ross Millikan, Jul 08 '15 at 05:08

score 2 · Answer 1 · edited Apr 13 '17 at 12:21

2

The problem is that the statement $2 \ln a = \ln (a^2)$ only works for $a > 0$.

For insight on why this is, this post explores that.

edited Apr 13 '17 at 12:21

Community

answered Jul 08 '15 at 04:02

MT_

thank, but If you could delete you're answer, because I want to delete the question please – Leman Jul 08 '15 at 04:09

1 Answers1