Negative-base logarithm, where's the issue here

Question

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

Definitions

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.

Question

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?

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.