# Can you take the natural log of a negative number?

##### 2 Answers

#### Answer:

If you are only working in real numbers, no, you cannot take the natural log of a negative number. If you are working in complex (imaginary) numbers, however, it is possible. Explanation below.

#### Explanation:

Notice that

#ln(-1) = pi i# .

Combining this with

#ln(-x) = ln(x) + pi i# when#x > 0#

However, notice that with this definition:

#ln(-1) + ln(-1) = ln(1) + pi i + ln(1) + pi i = 2 pi i#

Whereas:

#ln(-1 * -1) = ln(1) = 0#

So the

So what's going on?

The problem is that while

So

We can define a *principal* Complex logarithm consistent with the most common definition of

#ln(r(cos theta + i sin theta)) = ln(r) + i theta#

where

This principal Complex logarithm satisfies:

#e^(ln(z)) = z# for any#z in CC "\" {0}#

#ln(e^z) = z + 2pi k i# for some integer#k# for any#z in CC#

#### Answer:

If you are restricted to Real numbers, then "no".

If you are working with Complex numbers see George C.'s response.

#### Explanation:

[I've included this answer only because of uncertainty of the mathematics level at which the question was asked.]

If you could take the (natural) log of a negative number (say

then

would mean

Since