mathmath333
  • mathmath333
soft question
Mathematics
  • Stacey Warren - Expert brainly.com
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jamiebookeater
  • jamiebookeater
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mathmath333
  • mathmath333
if \(y=\log x,\ \{x,y\}\in \mathbb{R}\) can \(x\) be \(0\) or \(x<0\)
xapproachesinfinity
  • xapproachesinfinity
no! e^y=0???? can you find such number that goes with that
xapproachesinfinity
  • xapproachesinfinity
or any base actually b^a=0!!

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More answers

Michele_Laino
  • Michele_Laino
x can be only >0
mathmath333
  • mathmath333
and what about \(x<0\)
xapproachesinfinity
  • xapproachesinfinity
the same reason goes for x<0 e^y is always positive thus log x for x>0
Michele_Laino
  • Michele_Laino
the logarithm function is defined for positive numbers only, so x can not be < 0
xapproachesinfinity
  • xapproachesinfinity
this restrictions are there for the bond to exp
mathmath333
  • mathmath333
ok thanks
Michele_Laino
  • Michele_Laino
:)
mathmath333
  • mathmath333
by the way is there a logarithm for negative numbers
xapproachesinfinity
  • xapproachesinfinity
there is only one possibility that a power result in zero that is the case 0^0 but that is a calculus problem hehe
xapproachesinfinity
  • xapproachesinfinity
it is not quite zero but not 1 either!
Michele_Laino
  • Michele_Laino
no, there is not a logarithm of negative number
xapproachesinfinity
  • xapproachesinfinity
you mean log base negative ?
mathmath333
  • mathmath333
and what about complex numbers
mathmath333
  • mathmath333
i mean where x can be taken negative
xapproachesinfinity
  • xapproachesinfinity
no cannot x>0 always
Michele_Laino
  • Michele_Laino
yes! I think that the logarithm of a complex number is defined
xapproachesinfinity
  • xapproachesinfinity
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
xapproachesinfinity
  • xapproachesinfinity
but logs take multiple values in some way! if we allow it to be complex function
Michele_Laino
  • Michele_Laino
if we define a complex number like this: \[\Large z = \rho {e^{i\theta }}\] then the logarithmic function is: \[\Large f\left( z \right) = \ln z = \ln \rho + i\theta \]
Michele_Laino
  • Michele_Laino
where the subsequent additional contition holds: \[\Large 0 \leqslant \theta < 2\pi \]
xapproachesinfinity
  • xapproachesinfinity
it is a whole other interesting place :) complex numbers tend to solve such problem with some good tricks
Michele_Laino
  • Michele_Laino
that's right! @xapproachesinfinity

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