MHB Understanding Squaring the Bottom of an Equation

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The discussion focuses on the mathematical concept of squaring the bottom of an equation, specifically in the context of logarithmic properties. The user initially questions why the bottom is squared when dealing with negative values, such as -2. The explanation provided clarifies that the transformation follows the rule of logarithms, where -2 ln|x + 1/2| equates to ln(1/(x + 1/2)²). The conversation concludes with the user expressing understanding of the exponent law applied in this context. Overall, the thread emphasizes the importance of logarithmic identities in simplifying expressions.
Petrus
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Hello MHB,
I have problem understanding the last part, why do they square the bottom?
j5b5z9.png


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

Regards,
$$|\pi\rangle$$
 
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Petrus said:
Hello MHB,
I have problem understanding the last part, why do they square the bottom?
j5b5z9.png


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

Regards,
$$|\pi\rangle$$

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

$\chi$ $\sigma$
 
chisigma said:
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.
ea0d010db0bb2795fe2a83ea998cbd9c.png

right?

Regards,
$$|\pi\rangle$$
 
Petrus said:
Ohh now I see. We use this rule.
ea0d010db0bb2795fe2a83ea998cbd9c.png

right?

Regards,
$$|\pi\rangle$$

Right.

It also uses the exponent law $a^{-b} = \dfrac{1}{a^b}$
 
Thanks for the fast responed and help from you both!:)Now I understand!:)

Regards,
$$|\pi\rangle$$
 
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