Can you take the natural log of a negative number?

2 Answers
Jan 25, 2016

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 #e^(i pi) = -1# so in some sense you might like to define:

#ln(-1) = pi i#.

Combining this with #ln(ab) = ln(a) + ln(b)# you might then define:

#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 #ln(ab) = ln(a) + ln(b)# rule does not quite work for Complex logarithms.

So what's going on?

The problem is that while #e^x:RR->(0,oo)# is a one-one function as a Real valued function, the function #e^z:CC -> CC "\" {0}# is many to one.

So #e^z# does not have a well defined inverse #ln(z):CC "\" {0}->CC# unless we restrict the domain of #e^z# or the range of #ln(z)#.

We can define a principal Complex logarithm consistent with the most common definition of #Arg(z) in (-pi, pi]# as follows:

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

where #r > 0# and #theta in (-pi, pi]#

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#

Feb 2, 2016

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 #m<0#)
then
#color(white)("XXX")lnm=k#
would mean
#color(white)("XXX")e^k=m < 0#

Since #e~~2.7>0#
#color(white)("XXX")e^k > 0# for all Real values of #k#