Understanding Squaring the Bottom of an Equation

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SUMMARY

The discussion clarifies the mathematical principle of squaring the bottom of an equation in logarithmic expressions. Specifically, the transformation of $$-2\ \ln |x+\frac{1}{2}|$$ into $$\ln \frac{1}{(x+\frac{1}{2})^{2}}$$ is confirmed through the application of logarithmic rules. The participants emphasize the exponent law $$a^{-b} = \dfrac{1}{a^b}$$ as a foundational concept in this context. This understanding resolves the initial confusion regarding the necessity of squaring the denominator based on the logarithmic properties.

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  • Basic knowledge of algebraic manipulation involving fractions and exponents.
  • Ability to interpret mathematical notation and expressions accurately.
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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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