Okay, let's see. The problem is about finding the domain of the function log(x²) and then determining a + b where the domain is given as x < a or x > b. Hmm. Alright, let me start by recalling what the domain of a logarithmic function is. The logarithm function log(y) is defined only when y is positive. So, for log(x²), the argument inside the log, which is x², must be greater than zero. 

Wait, x² is always non-negative, right? Because any real number squared is positive or zero. So, x² ≥ 0. But since log is only defined for positive arguments, x² must be greater than zero. Therefore, x² > 0. 

So, solving x² > 0. When is x² greater than zero? Well, x² is zero only when x is zero. So, x² > 0 implies that x ≠ 0. Therefore, the domain of log(x²) is all real numbers except x = 0. 

But the problem states that the domain is x < a or x > b. So, this must mean that the domain is split into two intervals: one from negative infinity to a, and the other from b to positive infinity. But according to my previous conclusion, the domain is all real numbers except zero. Therefore, the domain should be x < 0 or x > 0. 

Wait, but the problem says the domain is x < a or x > b. So, comparing this with x < 0 or x > 0, it seems that a and b must be 0. Because if a is 0, then x < 0, and if b is 0, then x > 0. So, a + b would be 0 + 0 = 0. 

But let me double-check. Maybe I made a mistake here. Let's verify again. The function is log(x²). The argument x² must be positive, so x² > 0. Which is equivalent to x ≠ 0. Therefore, the domain is all real numbers except x = 0. So, in interval notation, that's (-∞, 0) ∪ (0, ∞). 

But the problem states the domain is x < a or x > b. So, comparing the two, it's (-∞, a) ∪ (b, ∞). So, to match with (-∞, 0) ∪ (0, ∞), we must have a = 0 and b = 0. Therefore, a + b = 0 + 0 = 0. 

Wait, but maybe the problem is written in a way that a and b are different. Let me think again. Suppose someone might interpret log(x²) as log(x)^2, but no, the notation log x² is standard for log(x²), not (log x)^2. So, the domain is still x ≠ 0. 

Alternatively, maybe there's a different approach. Let me consider the function log(x²). Let's analyze it step by step. 

First, log(x²) can be rewritten using logarithm properties. Remember that log(a^b) = b log(a). So, log(x²) = 2 log|x|. Wait, that's interesting. So, log(x²) is equal to 2 log|x|. Therefore, the function is equivalent to 2 log|x|. 

Now, the domain of log|x| is |x| > 0, which again is x ≠ 0. So, the domain is the same as before, x ≠ 0. Therefore, the domain is (-∞, 0) ∪ (0, ∞). 

So, in the problem statement, the domain is given as x < a or x > b. Comparing, a must be 0 and b must be 0. Therefore, a + b = 0. 

But let me check if there's a possibility that the problem is considering a different base or something else. Wait, the problem just mentions log x², without specifying the base. But regardless of the base, the domain would still be the same. Because logarithmic functions are defined when their arguments are positive. So, even if the base is 10 or e, the domain is still x² > 0, which is x ≠ 0. 

Therefore, the answer should be 0. 

Wait, but maybe the problem is written in a way that the user is supposed to find a and b such that the domain is x < a or x > b, but perhaps a and b are different? Let me think. Suppose the function was log(x)², then the domain would be x > 0, but that's not the case here. The problem is log(x²). 

Alternatively, maybe the problem is written with a typo, and the function is log(x) + log(x). But that would be 2 log x, which would have domain x > 0. But the problem states the domain is two intervals, so that can't be. 

Alternatively, maybe the problem is log(x^2) where x^2 is the argument. Which is the same as before. So, I think my initial conclusion is correct. 

Therefore, the answer is 0. 

But just to make sure, let's consider another approach. Let's suppose that someone might think that log(x^2) is defined when x is not zero, but maybe they consider the square root or something else. Wait, no. The square of x is always positive unless x is zero. So, the argument of the logarithm must be positive, so x can't be zero. 

Therefore, the domain is all real numbers except zero, which is written as x < 0 or x > 0. So, a is 0 and b is 0, so a + b is 0. 

I think that's solid. There's no other way around it. The domain is x ≠ 0, so the intervals are (-∞, 0) and (0, ∞). Therefore, a and b are both 0, so their sum is 0. 

Final Answer: \boxed{0}
Therefore, the final answer is $\boxed{0}$.
</think>