What is the logarithm of a negative number?

1 Answer
Aug 2, 2014

Logarithms of negative numbers are not defined in the real numbers, in the same way that square roots of negative numbers aren't defined in the real numbers. If you are expected to find the log of a negative number, an answer of "undefined" is sufficient in most cases.

It is possible to evaluate one, however, the answer will be a complex number. (a number of the form #a + bi#, where #i = sqrt(-1)#)

If you're familiar with complex numbers and feel comfortable working with them, then read on.

First, let's start with a general case:

#log_b (-x) = ?#

We will use the change-of-base rule and convert to natural logarithms, to make things easier later:

#log_b(-x) = ln(-x)/lnb#

Note that #ln(-x)# is the same thing as #ln(-1 * x)#. We can exploit the addition property of logarithms, and separate this part into two separate logs:

#log_b(-x) = (lnx + ln(-1))/lnb#

Now the only problem is figuring out what #ln(-1)# is. It might look like an impossible thing to evaluate at first, but there is a pretty famous equation known as Euler's Identity that can help us.

Euler's Identity states:

#e^(ipi) = -1#

This result comes from power series expansions of sine and cosine. (I won't explain that too in-depth, but if you are interested, there is a nice page here which explains a bit more)

For now, let us simply take the natural log of both sides of Euler's Identity:

#ln e^(ipi) = ln(-1)#


#ipi = ln(-1)#

So, now that we know what #ln(-1)# is, we can substitute back into our equation:

#log_b(-x) = (lnx + ipi)/lnb#

Now you have a formula for finding logs of negative numbers. So, if we want to evaluate something like #log_2 10#, we can simply plug in a few values:

#log_2(-10) = (ln10 + ipi)/ln2#

#approx 3.3219 + 4.5324i#