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The Log of a Negative Number

Date: 03/27/98 at 12:37:55
From: Brett Witty
Subject: Complex powers


I ran into a revision problem with logs which wasn't the problem 
itself, just a gaping hole in my learning. They had given us what 
turned out to be a log of a quadratic, with a x-3 factor or something. 
I was checking my answers like the good student I was, but my 
calculator couldn't do log (-3). I guessed that the answer had to be 
complex. I consulted my Maths coordinator and he agreed with me and 
said that (for example), log -10 equalled i. I had a play around at 
home and devised that x^i = -x (where x is a real number) and a few 
other rules for complex numbers for powers. Am I right? I tried to 
look up several maths databases for verification, but they didn't have 

I congratulate you all for all the helpful info and hard work you guys 
put into the site. It's great help.


Brett W

Date: 04/01/98 at 09:44:56
From: Doctor Barrus
Subject: Re: Complex powers

Hi, Brett!

You're partially right. To start out with, the logarithm of a negative 
number in base 10 is complex.

I ran into a problem in verifying what your coordinator said, though. 
Here's what I did, using the math software program Maple V (release 

(In the following, I'll use log[a](b) to mean the logarithm with base 
a of b. For example, log[2](8) = 3, since 2^3 = 8.)

First I used the base-conversion rule of logarithms:

     log[a](b) = log[c](b) / log[c](a)

to write:

     log[10](-10) = log[e](-10) / log[e](10) = ln(-10) / ln(10) 

(I did this because Maple can only work with complex logs when the 
logarithm base is e.)

Next I found the value of ln(-10):

     ln(-10) = ln(10 * -1)         
             = ln(10) + ln(-1)     (Using the sum rule of logs)

To deal with the ln(-1), I used Euler's Identity:

     e^(i*x) = cos(x) + i*sin(x)

In particular, e^(i*pi) = cos(pi) + i*sin(pi) = -1 + i*0 = -1

(I'm not sure how much math you've learned. The Euler identity is 
something that you can derive using Taylor series, which you learn 
about in calculus. Also, if you haven't covered trigonometry yet, 
you'll have to take my word for it that when you're measuring angles 
in radians (not degrees), cos(pi) = -1 and sin(pi) = 0).

Now, since e^(i*pi) = -1, we can write:

          -1 = e^(i*pi))
      ln(-1) = ln(e^(i*pi))
      ln(-1) = i*pi

So, back to our equation:

     ln(-10) = ln(10) + ln(-1)
             = ln(10) + (i*pi)
             = 2.303 + 3.1412*i (about)

which is the numerical answer Maple gives. Now ln(10) is about 2.303, 
so (going back to our first formula):

     log[10](-10) = ln(-10)/ln(10)

                    ln(10) + pi*i
                  = ---------------

                  = 1 + [pi/ln(10)]*i

                  = 1 + 1.364*i (about)
Now, raising 10 to this power gives -10, but this is not equal to i. 
So log(-10) is not equal to i.

Using Maple, I calculated 10^i and got -0.668 + 0.744*i 
(approximately), which, as you can see, is not equal to -10.

So your rule isn't quite correct, but you're right about the logs of 
negative numbers being complex. Using the steps I used to find out 
log(-10), you can find out log[a](-a) for any a. You might give it a 
shot. Good luck!

-Doctor Barrus,  The Math Forum
Check out our web site!  http://mathforum.org/dr.math/   

Associated Topics:
High School Logs

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