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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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if \(y=\log x,\ \{x,y\}\in \mathbb{R}\) can \(x\) be \(0\) or \(x<0\)
no! e^y=0???? can you find such number that goes with that
or any base actually b^a=0!!

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

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