Here we go again, basics of the basics. Faced with the following question.


Logarithm base a of x is by definition a number such as:

$$a^{\log_a x} = x$$

i.e. that answers the question "what power do I have to raise a in order to get x". I've read this answer about why logarithms can't be negative-based but that leaves out the case of "well-behaved" negative numbers.

$$(-2)^{3} = -8$$

so it is natural to expect that using the definition above and setting a = -2 and x = -8 we can define a logarithm in this case:

$$\log_{-2} (-8) = 3$$

which breaks the rule that requires the base of the logarithm to be positive and not 1, but sort of makes sense.


Following the rule of the base change, we can now do the following:

$$3 = \log_{-2} (-8) = {\log_2 (-8)\over \log_2 (-2)}$$

and now, that definitely makes no sense because on the left side we have something that's well-defined and on the right side we have something that's not defined at all, because there is no power that 2 can be raised to to yield -2 or -8. So my question is - where is the mistake? Which definition went wrong?


1 Answer 1


Definitions aren't right or wrong; claims about their consequences are. In this case, you've chosen $a,\,b,\,c$ so that $\log_ab=\frac{\log_ca}{\log_cb}$ breaks down; indeed, so will the proof of it.

The tricky thing about negative-base logarithms is that, unless you're prepared for them to be complex-valued, the set of values they can take isn't continuous. For example, what's $\log_{-2}3$? Well unfortunately, no real $x$ solves $(-2)^x=3$; in fact, $x,\,(-2)^{x}$ can only both be real if $x$ is a rational number which, in its lowest terms, has an odd denominator.

  • $\begingroup$ So, to reiterate: we don't want negative log bases if we aren't in a complex world because on a real number line such a log will no longer be a nice function? (nice = continuous and what not) $\endgroup$
    – Alma Do
    Jun 13, 2020 at 20:13
  • 1
    $\begingroup$ @AlmaDo Worse still, the set of values on which it's defined (with a real value) wouldn't include any intervals of positive width. $\endgroup$
    – J.G.
    Jun 13, 2020 at 20:14
  • $\begingroup$ Great!! This last comment answers my question (and I'm all fine with the log being "not well behaved" on a negative number line half so long as it is consistent with it being properly defined on some points on it) $\endgroup$
    – Alma Do
    Jun 13, 2020 at 20:55

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .