Quick Log Rule Confusion

Question

I want to simplify $$\log_{c}{-c}$$

I know that $$\log_{c}{c^k} = k$$

But is there a $\log$ rule for the first one?

Answer 1

Logarithms of negative numbers aren't defined on the set of real numbers $\mathbb{R}$. Note that for any positive number $c$ there is no $k\in\mathbb{R}$ such that $c^k=-c$. Also if you look at a graph of $y=\log x$ you can see that it's undefined for $x\leq0$.

Answer 2

$$\log_c(-c)=\log_c(-1*c)=\log_c(-1)+\log_c(c)=\log_c(-1)+1=\frac{\ln(-1)}{\ln(c)}+1=\frac{(2n-1)\pi i}{\ln(c)}+1$$ $$=(2n-1)\pi i\log_c(e)+1$$$$n=\pm0,\pm1,\pm2,...$$