Proof that $\log_b(-1)=0$

Question

So I made a proof that $\log_b(-1)=0$ using $\log_b(a^c)=c\log_b(a)$. Then, I did the following: $$\log_b(1)=\log_b((-1)^2)=2\log_b(-1)=0$$so$$\log_b(-1)=0$$What's wrong with this manipulation? We know that $\log_b(1)=0$ and $1=(-1)^2$. I am expecting this has to do with the different branches of the logarithm (which happens when a logarithm is in a false proof usually).

Answer 1

It's simply false in general that $\log_b(a^c)=c\log_b(a)$. That's for exactly the same reasons that $b^{ac}=(b^a)^c$ is false in general. You didn't 'make a proof' because you never proved the equation $\log(a^c)=c\log(a)$ (and you can't).

For no complex logarithm with $\log(1)=0$ will you also obtain $\log((-1)^2)=2\log(-1)$.

For an example where $\log((-1)^2)=2\log(-1)$ is true, consider the logarithm $\Bbb C^\star\to\Bbb C$ induced by $\pi\le\arg<3\pi$. There, $\log(-1)=\pi i$ and $\log(1)=2\pi i$.