Why isn't $2\log(-1)$ real? In high school we learn that a $a\log[(x)] = \log (x^a)$
From this I would assume $2\log(-1) = \log [(-1)^2]$
However, the first is not real and the second is, according to my calculator and textbook. Why is this?
 A: The formula $a\log x=\log x^a$ requires that $\log x$ exists. Here you want $\log(-1)$; this would a number such that $e^{\log(-1)}=-1$. But it turns out that the exponential of real numbers is always positive, so to write $\log(-1)$ you somehow need to extend the log function to the complex plane. 
Notice that if you keep going from your desired equality, you get 
$$
2\log(-1)=\log(-1)^2=\log 1 = 0,
$$
which would imply that $\log(-1)=0$. What this shows is that the property $a\log x=\log x^a$ does not hold for arbitrary complex numbers when you extend the log function to the complex plane. 
Regarding your first question, $\log(-1)$ cannot be real because the exponential of a real number is always positive. 
A: Logs are tricky when you start to discuss negative numbers.  To do this properly, you must concern yourself with branches of the log in the complex plane. Note that $e^{i\pi} = -1$ and that the exponential function has period $2\pi i$ in the complex plance.  
A: If the square were already inside the parenthesis like $\log((-1)^2)$ then you would indeed have $0$ as the result.
Beginning with the exponent outside, it is already the case that $\log(-1)$ must be evaluated outside the real numbers, in which case we would use the complex form $\log|a|+i\arg(a)$ which evaluates (at $-1$) to $0+i(2n+1)\pi$. Then the final expression is $2i(2n+1)\pi$.
