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