# Is Logarithm of negative real numbers defined? If so kindly explain.?

To the best of knowledge logarithm of negative real numbers are not defined. However logI-xI is defined. But, 2 contributors plus six voters say differently in the question http://answers.yahoo.com/question/index;_ylt=AsNSn...

As such I am now confused over this. I request contributors to kindly clarify on this.

Thanks for the contributors. But my interest is with respect to the question, where it is not stated that it is a complex number. It is only given as'y = -log₅(-x)' and asker desires 'Inverse of this function; When log is not defined, how we can state inverse?

### 6 Answers

- BenLv 610 years agoFavorite Answer
If we want to restrict to real numbers (which I think is natural when speaking of logarithms; as we've already seen in just two posts here, the complex exponential is not injective, so the inverse requires a restriction of domain [just like inverse trig functions]). Oops, lost track of that sentence :P

If we want to restrict to real numbers, then perhaps it will be helpful to keep track of domains and ranges of the functions concerned. (I know some mathematicians who don't like this terminology: a function's domain is part of its definition, not a consequence. So technically perhaps I should say we are looking for a "maximal domain of interest" or something.) First of all, the logarithm has domain (0,∞) and range R (reals). Notice that here we are saying that the logarithm of negatives is really undefined. Then its inverse (it has one since it is injective) must have domain R and range (0,∞)...which of course it does, as the inverse is just the standard exponential function.

Now, the function f(x)=-log_5(-x) has maximal domain (-∞,0), because for "log_5(stuff)" to make sense, "stuff" must be in (0,∞). So the inverse of this function will have domain R, and range (-∞,0). Notice that in the second link, this is in fact true.

- 10 years ago
Well!!!!!!!

Yes... They are defined

But, they are no longer real!!

By Euler's identity, e^(i π) + 1 = 0 .................. {1}

==> e^(i π) = - 1

Taking natural logarithm on both sides, we get i π = ln (-1) ................{2}

Thus, the negative numbers constitute the complex part of the logarithms!!

Suppose we define the complex number z = a e^(i b)

such that a and b are real numbers...

Then, ln z = ln [a e^(i b)] = ln a + ln [e^(i b)] = ln a + i b

==> ln z = ln a + ib

This is applicable to logarithms of all bases!!

The logarithms can be calculated for any bases by using the logarithmic rules of manipulation

For instance, if we define 2^x = -1, x can be definitely found!!

x = log₂ (-1) = ln (-1) . log₂ e = i π * 1.442 = 4.53 i

P.S: If we are free to enter the regime of complex numbers, logarithms are defined for any values and any bases!!

Similar to negative Numbers, Logarithms are defined for negative bases too!!!

THen, Logarithms are defined for any real number except for the base 0 and 1

Source(s): https://ccrma.stanford.edu/~jos/st/Logarithms_Nega... http://mathforum.org/library/drmath/view/55592.htm... - 4 years ago
In view that (-2)(-2)(-2) = -8, that you can say that logbase-2(-eight) = three, however what will you do then about constructive numbers? You run into the identical crisis. An extra, distinct, challenge arises with the exponential definition of logs of negatives numbers. Whilst e^iπ = -1, so is e^-iπ, and, indeed, e^iπ(2n+1), so that you become with an ambiguous multivalued operate ln(-1) = iπ(2n + 1) has an unlimited quantity of values.

- 10 years ago
Logs are a way of simplifying exponents. By there nature, we lose the ability to take the log of a negative number. It is the cost of simplifying.

But I have never found this a hardship because in the real world we can usually get away with changing our frame of reference to avoid the conflict.

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- Anonymous10 years ago
They are defined, but they are complex numbers. This follows from if:

z = r(cos θ + i sin θ) = r*e^(iθ),

then:

ln(z) = ln(r) + θi.

Then, if z is explicitly real and negative:

z = -r and θ = π.

This yields:

ln(-r) = ln(r) + πi.

You can read more here:

http://en.wikipedia.org/wiki/Complex_logarithm

I hope this helps!

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Yeah, I made a stupid mistake. =D

The reason why ln(-r) = ln(r) - πi works too is because logarithms are periodic with a period of 2πi. Hence, ln(-r) = ln(r) + πi = ln(r) - πi.