Am I right in thinking that #ln(-6)# is undefined ?

3 Answers
Aug 31, 2017

#"correct"#

Explanation:

#"the log function is not defined for negative values"#

#"the graph of ln should confirm this for you"#
graph{lnx [-10, 10, -5, 5]}

Aug 31, 2017

It depends...

Explanation:

As a real valued function of real numbers the function #e^x# is a one to one function from #(-oo, oo)# onto #(0, oo)#.

Therefore it has a well defined inverse function #ln(x)# from #(0, oo)# onto #(-oo, oo)#.

This real valued logarithm is not defined for any real number in #(-oo, 0]#

However, #e^x# can also be considered as a complex valued function of complex numbers.

As such, it is a many to one function from #CC# onto #CC"\"{0}#.

Since it is many to one, the complex valued logarithm #ln x# that extends the real valued logarithm has multiple branches. The principal logarithm is usually taken to have values in:

#{ x+yi : x in RR, y in (-pi, pi] }#

If #x# is any negative real number, then the principal value of its complex logarithm is:

#ln x = ln (-x) + pi i#

So, for example:

#ln(-6) = ln(6) + pi i#

In general:

#ln(r(cos theta + i sin theta)) = ln r + theta i" "# for #theta in (-pi, pi]#

Sep 1, 2017

Yes

Explanation:

Say you have the following expression:

#ln(-6) = x#

This directly translates, by the definition of the natural log, to

#e^x = -6#

#e# is a positive number equal to approximately 2.72. No matter how many times you multiply a positive number by itself, you will always still get a positive number, so #e^x# can never equal #-6#, and the answer is considered undefined.