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ksaimouli Group Title

slove

  • one year ago
  • one year ago

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  1. ksaimouli Group Title
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    |dw:1361407430942:dw|

    • one year ago
  2. ksaimouli Group Title
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    how this gets to \[\ln \sqrt{2}\]

    • one year ago
  3. ksaimouli Group Title
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    @TuringTest

    • one year ago
  4. tkhunny Group Title
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    ln(1) = 0

    • one year ago
  5. ksaimouli Group Title
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    i know that so |dw:1361408007302:dw|

    • one year ago
  6. tkhunny Group Title
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    You MUST have been given fundamental principles of logarithms, Here's one: log(a) + log(b) = log(a*b) Have you any others?

    • one year ago
  7. ksaimouli Group Title
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    i dont get this

    • one year ago
  8. ksaimouli Group Title
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    this is - so division right

    • one year ago
  9. tkhunny Group Title
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    Good. log(a) - log(b) = log(a/b) Any more?

    • one year ago
  10. ksaimouli Group Title
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    |dw:1361408413459:dw|

    • one year ago
  11. ksaimouli Group Title
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    so ln(2)/(2)

    • one year ago
  12. ksaimouli Group Title
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    how did they got ln sqrt 2

    • one year ago
  13. ksaimouli Group Title
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    @SithsAndGiggles

    • one year ago
  14. tkhunny Group Title
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    So, you're saying you have only two properties for logarithms? You don't have this one: \(a\cdot \log(b) = \log\left(b^{a}\right)\)?

    • one year ago
  15. ksaimouli Group Title
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    |dw:1361408835272:dw| where do u use that

    • one year ago
  16. tkhunny Group Title
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    \(\dfrac{\ln(2)}{2} = \dfrac{1}{2}\cdot \ln(2) = ln\left(2^{1/2}\right) = \ln(\sqrt{2})\) It's all in the properties!

    • one year ago
  17. ksaimouli Group Title
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    wow thx

    • one year ago
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