Logarithm rules for complex numbersAre the logarithm rules true for complex numbers?We know that for...

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 $z\ln (b)-d\ln (a)=2\pi in$ for some integer $n$. So the result is
${\log }_b(a^d)={\displaystyle \frac{d\ln (a)+2\pi in}{\ln (b)}}$
Similarly,
${\log }_b(a)={\displaystyle \frac{\ln (a)+2\pi im}{\ln (b)}}$
for some integer $m$. And thus (assuming you use the same values of $\ln (a)$ and $\ln (b)$ in both cases)
${\log }_b(a^d)-d{\log }_b(a)=2\pi i{\displaystyle \frac{n-md}{\ln (b)}}$
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).