Logarithm rules for complex numbers Are the logarithm rules true for complex numbers? 
We know that for positive real numbers $a$, $b$, $c$ and real number $d$ that:
$$\log_b\left(a^d\right)=d\log_b(a)$$
$$\log_b(a) = \frac{\log_c(a)}{\log_c(b)}$$
$$\log_b(xy)=\log_b(x)+\log_b(y)$$
$$\log_b\left(\frac xy\right)=\log_b(x)-\log_b(y)$$
We also know that 
$$\log_b\left(b^d\right)=d$$
Does this extend to complex numbers as $a$, $b$, $c$, or $d$?

My instinct is that
$$\log_b\left(b^{s+t\mathrm i}\right) = s+t\mathrm i$$
In other words, I'm pretty confident that the last formula works when $d$ is a complex number
 A: You have to be careful because logs and non-integer powers are multivalued functions.  The definition is that $a^d = \exp(d \ln(a))$ (for any branch of $\ln$).  Now $\log_b(a^d)$ is any $z$ such that $b^z = a^d$, i.e.
$\exp(z \ln(b)) = \exp(d \ln(a))$, and that is equivalent to
$z \ln(b) - d \ln(a) = 2 \pi i n$ for some integer $n$. 
 So the result is
$$ \log_b(a^d) = \dfrac{d \ln(a) + 2 \pi i n}{\ln(b)}$$
Similarly, $$\log_b(a) = \dfrac{\ln(a) + 2 \pi i m}{\ln(b)}$$
for some integer $m$.  And thus (assuming you use the same 
values of $\ln(a)$ and $\ln(b)$ in both cases)
$$\log_b(a^d) - d \log_b(a) = 2 \pi i \dfrac{n - m d}{\ln(b)}$$
For example, take $a = b = e$ and $d = 2 \pi i$, and use the principal
branch of $\ln$.  $\log_e(e^{2\pi i}) = \ln(1) = 0$ but $2 \pi i \log_e(e) = 2 \pi i$.
Another interesting example is $a = b = -1$, $d = 3$.  Now $(-1)^3 = -1$, 
but there is no way to have $\log_{-1}(-1) = \log_{-1}((-1)^3) = 3 \log_{-1}(-1)$ (this would imply $\log_{-1}(-1) = 0$, which is certainly false).
