Show $(\ln(x^2))^2-(\ln x)^2=3(\ln x)^2$

Question

I read an example on integrals.

I can't see how $$(\ln(x^2))^2-(\ln x)^2=3(\ln x)^2.$$

It's only true for $x>0$, but otherwise all answers below are correct. — JPi, Apr 01 '14 at 16:40

score 4 · Answer 1 · edited Apr 01 '14 at 16:42

4

Simply use the following equation:

$$\ln (x^2) = 2 \ln (x)$$

edited Apr 01 '14 at 16:42

JPi

answered Apr 01 '14 at 16:10

Peter

score 0 · Answer 2 · answered Apr 01 '14 at 16:14

0

$$a^2-b^2=(a+b)(a-b)$$

Therefore

$$(\ln(x^2))^2-(\ln(x))^2=(\ln(x^2)+\ln(x))(\ln(x^2)-\ln(x))$$

Also we have,

$$\ln(x^2)=2\ln(x)$$

answered Apr 01 '14 at 16:14

Guy

score 0 · Answer 3 · answered Apr 01 '14 at 16:14

0

Alternatively,

$$(\ln x^2)^2 - (\ln x)^2 = (\ln x^2 + \ln x)(\ln x^2 - \ln x)\\ = \ln x^3 \ln x\\ = 3 \ln x \ln x\\ = 3 (\ln x)^2$$

answered Apr 01 '14 at 16:14

Yiyuan Lee

score 0 · Answer 4 · edited Apr 01 '14 at 16:36

0

$(\ln(x^2))^2−(\ln x)^2=3(\ln x)^2$

using ln(x^2)=2ln(x)

(ln(x^2))^2−(lnx)^2 = 4(lnx)^2−(lnx)^2 = 3(lnx)^2

edited Apr 01 '14 at 16:36

Amzoti

answered Apr 01 '14 at 16:14

AnarKi

4 Answers4