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maheshmeghwal9

Plz help:)

  • one year ago
  • one year ago

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  1. dpaInc
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    hey mahesh

    • one year ago
  2. maheshmeghwal9
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    • one year ago
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  3. maheshmeghwal9
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    why b>1? Why not b>0?

    • one year ago
  4. maheshmeghwal9
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    in the attachment.

    • one year ago
  5. dpaInc
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    holymoly that's one big pic.

    • one year ago
  6. maheshmeghwal9
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    should I make it small?

    • one year ago
  7. dpaInc
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    sawrite....

    • one year ago
  8. maheshmeghwal9
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    • one year ago
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  9. dpaInc
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    b>1 because the negatitve exponents will take care of values for b between 0 and 1... that make sense?

    • one year ago
  10. dpaInc
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    so it is defined for b>1

    • one year ago
  11. maheshmeghwal9
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    didn't understand ya main concept:(

    • one year ago
  12. dpaInc
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    let's do this then...

    • one year ago
  13. dpaInc
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    |dw:1337931693406:dw|

    • one year ago
  14. dpaInc
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    do you see why b doesn't have to be defined between 0 and 1?

    • one year ago
  15. maheshmeghwal9
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    but what is the problem for 0<b<1.Isn't it defined for exponential f(x) too. Since it also gives a value which is defined as "x". in ur example|dw:1337932045573:dw| All r defined:)

    • one year ago
  16. dpaInc
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    there's no problem... with 0<b<1.. it's just not needed...

    • one year ago
  17. maheshmeghwal9
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    k! did it have no need due to log?

    • one year ago
  18. dpaInc
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    notice the base of logs CAN have a base between 0 and 1 also but could be transformed and written with a base bigger than 1

    • one year ago
  19. princeofwetlot
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    is there a reason why this was bumped?

    • one year ago
  20. maheshmeghwal9
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    Hi @experimentX I couldn't understand the my basic questiom.Plz help:)

    • one year ago
  21. experimentX
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    I don't understand the reason why b should be greater than 1, but b should always be greater than zero ... for that expression to be defined

    • one year ago
  22. maheshmeghwal9
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    ya I also think this:)

    • one year ago
  23. experimentX
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    \[ \log_b x = \frac{\log_k x}{\log_k b}\] Also between 0-1 \( \log_k b < 1\)

    • one year ago
  24. maheshmeghwal9
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    ok!

    • one year ago
  25. maheshmeghwal9
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    I had this problem while studying Logarithmic Differentiation. I think there is some connection between b>1 & Logarithmic Differentiation.

    • one year ago
  26. experimentX
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    \[ \frac{d}{dx} (\log_b x) = \frac{1}{x \ln b}\] At b=1, \( \ln b =0\) is zero ... so it is undefined, so is the above expression, but for (0,1) the expression is defined http://www.wolframalpha.com/input/?i=plot+y+%3D+d%2Fdx+log[0.5%2C+x]+from+0+to+10 http://www.wolframalpha.com/input/?i=plot+y+%3D+log[0.5%2C+x]+from+0+to+10

    • one year ago
  27. maheshmeghwal9
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    So indirectly do u want to say that b>1 is right there! Please tell:)

    • one year ago
  28. experimentX
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    No ... i think i got it wrong it somewhere check this out http://www.wolframalpha.com/input/?i=plot+y+%3D+log[x%2C+2]+from+0+to+10

    • one year ago
  29. maheshmeghwal9
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    k! thanx alot:)

    • one year ago
  30. experimentX
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    wait ...

    • one year ago
  31. maheshmeghwal9
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    k!

    • one year ago
  32. experimentX
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    it seems i was right ... but somewhat not accurate http://www.wolframalpha.com/input/?i=plot+y+%3D+log[x%2C+5]+from+0+to+2

    • one year ago
  33. maheshmeghwal9
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    K! I understood that.I also think u r right.

    • one year ago
  34. experimentX
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    at b=0, \( \log_b x\) is zero ... and \( dy/dx\) is indeterminate

    • one year ago
  35. maheshmeghwal9
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    ok! thanx.

    • one year ago
  36. maheshmeghwal9
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    ow I have some clues. I will think on that point more. Thanx for clues:)

    • one year ago
  37. maheshmeghwal9
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    ow=now

    • one year ago
  38. experimentX
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    http://www.wolframalpha.com/input/?i=plot+y+%3D+d%2Fdx%28log[b%2C+x]%29+from+b%3D0+to+2 check this out

    • one year ago
  39. maheshmeghwal9
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    Hmm! I checked that:)

    • one year ago
  40. experimentX
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    b cannot be 1 and less than zero ... that's all

    • one year ago
  41. maheshmeghwal9
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    k! but can be (0,1)

    • one year ago
  42. experimentX
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    i guessed so ...

    • one year ago
  43. maheshmeghwal9
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    so we again came there; where was our doubt:)

    • one year ago
  44. maheshmeghwal9
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    this is a very confusing state; isn't it?

    • one year ago
  45. experimentX
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    well yes!!

    • one year ago
  46. maheshmeghwal9
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    What should we do now? by the way I have asked this to my sir also online, but they will receive question & give answer after almost a week!!!

    • one year ago
  47. experimentX
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    what was the original question??

    • one year ago
  48. maheshmeghwal9
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    Pardon me what do u want?; that is a video cutted part. I had this problem while studying video:)

    • one year ago
  49. experimentX
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    Oh ... i thought this is a part of the problem where you got stuck!!

    • one year ago
  50. maheshmeghwal9
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    so is there anyone in OS who can solve our problem:)

    • one year ago
  51. experimentX
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    I don't know ...

    • one year ago
  52. maheshmeghwal9
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    k! np:) Thanx for ur co-operation. When my problem is solved, i'll tell u too.

    • one year ago
  53. experimentX
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    k

    • one year ago
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