$2\ln(-x) \neq \ln(-x^2) ? $

Question

I know the rule

$$n\ln(x) = \ln(x^n)$$

But this doesn't apply to $$2\ln(-x) = \ln(-x^2)$$

Can you see what I'm not understanding?

Watch your parentheses: $2\ln (-x) = \ln [(-x)^2]$ (which in turn equals $\ln (x^2)$). — Christopher Carl Heckman, Apr 18 '16 at 07:38
@CarlHeckman But when I type this into my calculator I get a different graph. $\ln(x^2)$ goes on beyond the positive $x$-axis where $2\ln(-x)$ does not. — Cro-Magnon, Apr 18 '16 at 07:41

5xum · Accepted Answer · 2016-04-18T07:46:59.013

3

The rule still applies, but you misused it. The rule is:

For every $a>0$ and $b\in\mathbb R$, $b\cdot \ln (a)$ is equal to $\ln(a^b).$

So, in this rule if you plug in $a=x$ and $b=n$, you get the rule you cite.

Now, what do you get if you plug in $b=2$ and $a=-x$?

edited Apr 18 '16 at 07:46

answered Apr 18 '16 at 07:42

5xum

Thank you so much. That explains it. – Cro-Magnon Apr 18 '16 at 07:42
@Cro-Magnon As always, if an answer is what you needed, you can upvote and accept it. – 5xum Apr 18 '16 at 07:43
@Joanpemo True, I fixed that. – 5xum Apr 18 '16 at 07:47

score 2 · Answer 2 · answered Apr 18 '16 at 07:44

The real logarithmic functions are defined only for $\;x>0\;$ , so you have to be careful with this restriction.

So we indeed have $\;\log(x^a)=a\log x\;$ for any $\;x>0\;$ , but if we have doubts we must use

$$\log((-x)^2)=2\log|x|$$

If $\;-x>0\implies 2\log(-x)=\log(-x)^2=\log x^2\;$, otherwise $\;-x<0\;$ and then $\;\log(-x)\;$ is not even defined.

2 Answers2