Understanding Squaring the Bottom of an Equation

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Discussion Overview

The discussion revolves around understanding the mathematical manipulation involving logarithms and exponents, specifically in the context of squaring terms in an equation. Participants explore the reasoning behind squaring the bottom of an expression and the implications of different coefficients in logarithmic identities.

Discussion Character

  • Technical explanation
  • Conceptual clarification
  • Mathematical reasoning

Main Points Raised

  • One participant expresses confusion about why the bottom of an equation is squared, questioning if the squaring is related to the coefficient of -2.
  • Another participant provides a mathematical identity involving logarithms, demonstrating that $$-2 \ln |x+\frac{1}{2}|$$ can be rewritten as $$\ln \frac{1}{(x+\frac{1}{2})^{2}}$$.
  • A later reply confirms the use of the logarithmic rule and introduces the exponent law $$a^{-b} = \frac{1}{a^b}$$ as part of the explanation.
  • Participants engage in a back-and-forth to clarify understanding, with one expressing newfound clarity after the explanation.

Areas of Agreement / Disagreement

Participants appear to reach a mutual understanding regarding the logarithmic manipulation, though initial confusion is evident. The discussion does not present any major disagreements, but rather a progression from confusion to clarity.

Contextual Notes

The discussion does not explicitly address any limitations or unresolved mathematical steps, focusing instead on the clarification of logarithmic properties.

Who May Find This Useful

Readers interested in logarithmic identities, mathematical manipulation of equations, or those seeking clarification on exponent rules may find this discussion beneficial.

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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