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mathmath333

  • one year ago

soft question

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  1. mathmath333
    • one year ago
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    if \(y=\log x,\ \{x,y\}\in \mathbb{R}\) can \(x\) be \(0\) or \(x<0\)

  2. xapproachesinfinity
    • one year ago
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    no! e^y=0???? can you find such number that goes with that

  3. xapproachesinfinity
    • one year ago
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    or any base actually b^a=0!!

  4. Michele_Laino
    • one year ago
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    x can be only >0

  5. mathmath333
    • one year ago
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    and what about \(x<0\)

  6. xapproachesinfinity
    • one year ago
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    the same reason goes for x<0 e^y is always positive thus log x for x>0

  7. Michele_Laino
    • one year ago
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    the logarithm function is defined for positive numbers only, so x can not be < 0

  8. xapproachesinfinity
    • one year ago
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    this restrictions are there for the bond to exp

  9. mathmath333
    • one year ago
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    ok thanks

  10. Michele_Laino
    • one year ago
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    :)

  11. mathmath333
    • one year ago
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    by the way is there a logarithm for negative numbers

  12. xapproachesinfinity
    • one year ago
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    there is only one possibility that a power result in zero that is the case 0^0 but that is a calculus problem hehe

  13. xapproachesinfinity
    • one year ago
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    it is not quite zero but not 1 either!

  14. Michele_Laino
    • one year ago
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    no, there is not a logarithm of negative number

  15. xapproachesinfinity
    • one year ago
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    you mean log base negative ?

  16. mathmath333
    • one year ago
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    and what about complex numbers

  17. mathmath333
    • one year ago
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    i mean where x can be taken negative

  18. xapproachesinfinity
    • one year ago
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    no cannot x>0 always

  19. Michele_Laino
    • one year ago
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    yes! I think that the logarithm of a complex number is defined

  20. xapproachesinfinity
    • one year ago
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    in complex theory, yes there are some stuff of that sort:) i didn't take complex analysis yet but i do believe they do some kind of tricks around that

  21. xapproachesinfinity
    • one year ago
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    but logs take multiple values in some way! if we allow it to be complex function

  22. Michele_Laino
    • one year ago
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    if we define a complex number like this: \[\Large z = \rho {e^{i\theta }}\] then the logarithmic function is: \[\Large f\left( z \right) = \ln z = \ln \rho + i\theta \]

  23. Michele_Laino
    • one year ago
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    where the subsequent additional contition holds: \[\Large 0 \leqslant \theta < 2\pi \]

  24. xapproachesinfinity
    • one year ago
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    it is a whole other interesting place :) complex numbers tend to solve such problem with some good tricks

  25. Michele_Laino
    • one year ago
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    that's right! @xapproachesinfinity

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