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Logarithm Law

Petrus

Well-known member
Feb 21, 2013
739
Hello MHB,
I have problem understanding the last part, why do they square the bottom?


Is it because we got -2? if we would have -3 would we take the bottom \(\displaystyle (bottom)^3\)?
I am aware that \(\displaystyle \ln|f(x)|- \ln|g(x)|= \ln\frac{f(x)}{g(x)}\)

Regards,
\(\displaystyle |\pi\rangle\)
 

chisigma

Well-known member
Feb 13, 2012
1,704
Hello MHB,
I have problem understanding the last part, why do they square the bottom?


Is it because we got -2? if we would have -3 would we take the bottom \(\displaystyle (bottom)^3\)?
I am aware that \(\displaystyle \ln|f(x)|- \ln|g(x)|= \ln\frac{f(x)}{g(x)}\)

Regards,
\(\displaystyle |\pi\rangle\)
Simply is...

$$- 2\ \ln |x+\frac{1}{2}| = \ln \frac{1}{|x+\frac{1}{2}|^{2}} = \ln \frac{1}{(x+\frac{1}{2})^{2}}$$

Kind regards

$\chi$ $\sigma$
 

Petrus

Well-known member
Feb 21, 2013
739
Simply is...

$$- 2\ \ln |x+\frac{1}{2}| = \ln \frac{1}{|x+\frac{1}{2}|^{2}} = \ln \frac{1}{(x+\frac{1}{2})^{2}}$$

Kind regards

$\chi$ $\sigma$
Ohh now I see. We use this rule.

right?

Regards,
\(\displaystyle |\pi\rangle\)
 

SuperSonic4

Well-known member
MHB Math Helper
Mar 1, 2012
249
Ohh now I see. We use this rule.

right?

Regards,
\(\displaystyle |\pi\rangle\)
Right.

It also uses the exponent law $a^{-b} = \dfrac{1}{a^b}$
 

Petrus

Well-known member
Feb 21, 2013
739
Thanks for the fast responed and help from you both!:)Now I understand!:)

Regards,
\(\displaystyle |\pi\rangle\)