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maheshmeghwal9

  • 3 years ago

Plz help:)

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  1. dpaInc
    • 3 years ago
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    hey mahesh

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

  4. maheshmeghwal9
    • 3 years ago
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    in the attachment.

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

  6. maheshmeghwal9
    • 3 years ago
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    should I make it small?

  7. dpaInc
    • 3 years ago
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    sawrite....

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

  10. dpaInc
    • 3 years ago
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    so it is defined for b>1

  11. maheshmeghwal9
    • 3 years ago
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    didn't understand ya main concept:(

  12. dpaInc
    • 3 years ago
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    let's do this then...

  13. dpaInc
    • 3 years ago
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    |dw:1337931693406:dw|

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

  15. maheshmeghwal9
    • 3 years ago
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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:)

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

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

  18. dpaInc
    • 3 years ago
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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

  19. princeofwetlot
    • 3 years ago
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    is there a reason why this was bumped?

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

  21. experimentX
    • 3 years ago
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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

  22. maheshmeghwal9
    • 3 years ago
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    ya I also think this:)

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

  24. maheshmeghwal9
    • 3 years ago
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    ok!

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

  26. experimentX
    • 3 years ago
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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

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

  28. experimentX
    • 3 years ago
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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

  29. maheshmeghwal9
    • 3 years ago
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    k! thanx alot:)

  30. experimentX
    • 3 years ago
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    wait ...

  31. maheshmeghwal9
    • 3 years ago
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    k!

  32. experimentX
    • 3 years ago
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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

  33. maheshmeghwal9
    • 3 years ago
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    K! I understood that.I also think u r right.

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

  35. maheshmeghwal9
    • 3 years ago
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    ok! thanx.

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

  37. maheshmeghwal9
    • 3 years ago
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    ow=now

  38. experimentX
    • 3 years ago
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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

  39. maheshmeghwal9
    • 3 years ago
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    Hmm! I checked that:)

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

  41. maheshmeghwal9
    • 3 years ago
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    k! but can be (0,1)

  42. experimentX
    • 3 years ago
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    i guessed so ...

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

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

  45. experimentX
    • 3 years ago
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    well yes!!

  46. maheshmeghwal9
    • 3 years ago
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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!!!

  47. experimentX
    • 3 years ago
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    what was the original question??

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

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

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

  51. experimentX
    • 3 years ago
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    I don't know ...

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

  53. experimentX
    • 3 years ago
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    k

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