Details of logarithmic calculations?

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Discussion Overview

The discussion revolves around the methods and calculations involved in determining logarithmic values, specifically focusing on how values like \(\log{2}\) and \(\log{3}\) are derived. Participants express interest in both the theoretical basis and practical calculations of logarithms.

Discussion Character

  • Exploratory
  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • Some participants inquire about the specific calculations for logarithmic values, such as \(\log{2}\) and \(\log{3}\).
  • One participant mentions that \(\log{2}\) is known to be 0.3010 and seeks clarification on how this value is calculated.
  • Another participant suggests using Taylor's Series as a method for calculating logarithmic values, noting that while the series converges slowly, it can eventually yield accurate results.

Areas of Agreement / Disagreement

Participants express curiosity about logarithmic calculations, but there is no consensus on the methods or specific calculations discussed. Multiple approaches and questions remain unresolved.

Contextual Notes

The discussion includes references to Taylor's Series for logarithmic calculations, but the convergence and accuracy of this method are not fully explored or agreed upon. Additionally, the assumptions behind the logarithmic values mentioned are not detailed.

kiru
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From where I can get the details of lograthmic calculations?How the values are found?and on what basis?
 
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Do you want to know for instance how [tex]\log{3}[/tex] is found? That kind of calculation?
 
Jameson said:
Do you want to know for instance how [tex]\log{3}[/tex] is found? That kind of calculation?
Actually I want to know this:We know that log{2} is 0.3010.How it is calculated?
 
kiru said:
Actually I want to know this:We know that log{2} is 0.3010.How it is calculated?

I would have the same question + how cosines are calculated.
 
There are a variety of ways.

The simplest is to use "Taylor's Series". It can be shown, in Calculus, that for x between 0 and 2, ln(x)= (x-1)- (1/2)(x-1)2+ (1/3)(x-1)3-...- ((-1)n/n)(x-1)n+... That infinite series does not converge very fast but it will give close to the correct value eventually.
 

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