Logarithm Law

Petrus

Well-known member
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
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
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
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
Thanks for the fast responed and help from you both!Now I understand!

Regards,
$$\displaystyle |\pi\rangle$$