Negative-base logarithm, where's the issue here
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.