Value of Logarithm of negative number Why the logarithmic value of negative number can't be define? Is there any special reason?
 A: The question is a good one and has motivated and prompted (and provided examples for) significant developments in understanding of complex (rather than real) functions.
Ponder, for example, with $r \in \mathbb Z, y \in \mathbb R$:$$e^{i\pi}=e^{3i\pi}=e^{(2r+1)i\pi}=-1$$
If $y=\ln x$ then
$$x=e^y=e^{y+2i\pi}=e^{y+2ri\pi}$$
$$-x=e^{y+i\pi}=e^{y+3i\pi}=e^{y+(2r+1)i\pi}$$
This shows that the logarithm as a real function does not tell the whole story, and as soon as you move to the complex numbers the situation becomes a whole lot more interesting.
A: In the real number system, it can't be defined because $a^x$ for positive reals $a$ and arbitrary reals $x$ is never negative. As the logarithm is the inverse of this operation (w.r.t. x), it does not exist for negative inputs.
A: There are some 'issues' with logarithms that may help you understand how negative arguments pose a problem.
To avoid misunderstandings about what we're discussing, let's first set out (and agree on) some basic notions:


*

*the BASE of the logarithm doesn't play a role in your question:
2Log(x) and 6Log(x), for instance, differ only by a scaling factor and any
considerations about negative arguments for 2log(x) apply equally to
6log(x). For the discussion, it's most convenient to pick the NATURAL
logarithm, which has base e; conventionally it is denoted ln(x) or sometimes log(x), i.e. without stating the 'base'.

*ln(x) is the inverse of the exponential function exp(y), i.e.
exp(ln(x)) = x and ln(exp(x)) = x.

*exp() and ln() take not only integer arguments, but all reals, and -
important here - also complex numbers. The function values themselves can also be complex-valued. Whereas the ln()-is-inverse-of-exp() is a run of the mill for real arguments, the generalization to complex numbers brings up some issues, as we will see shortly.

*A complex number c can be written as c = Re + i*Im, where Re and Im
are the real and imaginary parts respectively (each real-valued), and
i=sqrt(-1): the base unit along the imaginary axis
in the complex plane, in the same fashion as 1 is the base unit
along the real axis.

*The 'decomposition' "Re + i*Im" might be said to express c in cartesian coordinates, to give you the picture of a (complex) number 
as a point in a 2-dimensional space - the 'complex plane' - where the
axes are the real and imaginary axis and Re and Im are the coefficients of the "point" c along these axes.

*More convenient for our present discussion is to express c - still the same point in the 2-dimensional "complex plane" - in POLAR coordinates: c =
r*exp(i*a), where r is the distance from c to the origin, and a is
the angle that the vector (from origin to c) makes with the positive
x-axis (here: the Real axis), counting counter-clockwise from zero to
2*pi. Take a fixed value for r; then, increasing the angle a from
zero to 2*pi you describe a circle around the origin, with radius r, starting at the
positive x-axis, and proceeding counter-clockwise until we're back
again at where we started. (exp(i*a) can be written as cos(a) + i*sin(a). In fact, this is the way to switch back and forth between the 'cartesian' and the 'polar' forms of expressing a complex number.)


Having this out of the way, lets have a closer look at the remark about traversing a circle, corresponding to increasing the angle a. What happans if we increase the angle a further beyond 2*pi? No problem, we simply continue over the same circle as before - look at the 'cos(a) + i*sin(a)' expression to verify that you'll get the same values for a and for a+2*pi.
Here, we encounter a problem, however, when we bring in the fact that ln() and exp() are the inverse of each other. Let
y = exp(i*a) = exp(i*(a+2pi)) 

then, because ln() and exp() are the inverse of each other, we have:
ln(y) = ln(exp(i*a)) = i*a

but also
ln(y) = ln(exp(i*(a+2pi)) = i*(a+2pi)

and then, of course, ln(y) has an infinite number of values i*(a+k*pi), where k can be any positive or negative integer!
Obviously, it is not "handy" for using the logarithm in calculations to have it multi-valued as we've just demonstrated from its inverse-to-exp property. Leaving aside the interesting mathematics around this multi-valued aspect, we agree to DEFINE ln() - when we process it in calculations and formula's and so on - as SINGLE-valued. For this, we have to make a choice and the conventional choice is to let ln(i*a) only have values from -pi through +pi. Of course, it must be an 'interval' of 2*pi "width". Stating the same thing in another way, without any mentioning of logarithms: we agree that a point in the complex plane, if written in polar coordinates, has an angle a between -pi and + pi (although, again, the same point could be described by adding any multiple of 2pi to the angle).
The "choice" for the logarithm for its single-value value, is not quite arbitrary, by the way, and so not really 'just convention': we would be in an awkward position if the ln with complex arguments would not coincide with the ordinary ln() with real arguments
as soon as the complex argument happens to be 'just a real'. Indeed, we "know" that ln(e)=1 and not 1+2pi ,or 1+4pi, ...
We're not done yet with our logarithm. Consider again the circle in the complex plane, now happily confining the points on it to have an angle (called "the (principal) argument" of the point) between -pi and +pi, and hence the ln(x) for these points on the (r=1: unit-)circle to have values from -i*pi to +i*pi.
Focus on traversing that circle and approaching the NEGATIVE x-axis (real axis) from above, i.e. the angle being a bit less than pi and then converging to pi. When you arrive at the negative x-axis, your angle has become pi and the ln(x) has become i*pi.
Now, do the same thing from below: start with an angle a bit larger than -pi and let it decrease to -pi. When you arrive at the negative x-axis, your angle has become -pi and ln(x) has become -i*pi.
So, this multi-valued issue has not yet been totally eradicated: we have an ambiguity at the negative x-axis remaining. When you traverse the circle across the negative axis, your angle jumps from +pi to -pi, and ln(x) jumps from i*pi to -i*pi. What then, is the value at exactly the negative axis? We "solve" the question by "choosing" the positive variety, i.e. ln(-1)=+i*pi.
We arrive at two conclusions:
First, there is a discontinuity for the ln()-values at the negative x-axis when you cross that line. This holds for ALL the values on that axis (our discussion so far took always r=1, i.e. the unit-circle, but of course we may have any radius - smaller and larger circles - and always get the same problem). For this discontinuity, that negative x-axis is, for logarithms, called "the cut" in the complex plane.
Second, ln(x) DOES have a value for negative real x, i.e. for the points on the negative x-axis in the complex plane, but it's value (for r=1) is +i*pi, which is a complex number and not a real number. So, when you say that ln(x) is not defined for negative x, it would be more accurate to say that it's value is complex and therefore it has not a REAL value - which you might loosely - but less correctly - formulate as "not defined (... in the real-values world)".
Wikipedia has lots of mathematical pages, also on logarithms. For instance: http://en.wikipedia.org/wiki/Logarithm .
