I know that the logarithm function is only defined for positive arguments, so I need to find the values of $x$ that make $x^2$ positive. I can start by sketching the graph of $y = x^2$, which is a parabola that opens upward and has a minimum at $(0, 0)$. I want to exclude the values of $x$ that make $x^2$ zero or negative, which are the values between $-a$ and $a$, where $a$ is the positive root of $x^2 = 0$. I can use the zero product property to solve for $a$, and I get $a = 0$. So, the domain of $\log x^2$ is $x < 0$ or $x > 0$, which means that $a = 0$ and $b = 0$. Therefore, $a + b = 0 + 0 = 0$. 
If the domain of the function $\log x^2$ is $x < a$ or $x > b$, for some $a$ and $b$, find $a + b$.