Understanding Squaring the Bottom of an Equation

[Petrus]

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]

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]

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]

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]

Thanks for the fast responed and help from you both!:)Now I understand!:)

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