Why is $\log z$ continuous for $x\leq 0$ rather than $x\geq 0$? Explain why Log (principal branch) is only continuous on $$\mathbb{C} \setminus\{x + 0i: x\leq0\}$$ is the question.
However, I can't see why this is. Shouldn't it be $x \geq 0$ instead?
Thanks.

The important thing here is that the exponential function is periodic with period $2\pi i$, meaning that $e^{z+2\pi i}=e^z$ for all $z\in\Bbb C$.
If you imagine any stripe $S_a:=\{x+yi\mid y\in [a,a+2\pi)\}$ for any $a\in\Bbb R$, then we get that $z\mapsto e^z$ is actually one-to-one, restricted to $S_a$, (and maps onto $\Bbb C\setminus\{0\}$). A branch of $\log$ then, is basically the inverse of this $\exp|_{S_a}$ (more precisely, the inverse of $\exp|_{{\rm int\,}S_a}$). 
Observe that if a sequence $z_n\to x+ai\,$ and another $\,w_n\to x+(a+2\pi)i$ within the stripe $S_a$, then $\lim e^{z_n} = \lim e^{w_n}=:Z$. So, which value of the logarithm should be assigned to $Z$? Is it $x+ai$ on one edge of the stripe, or is it $x+(a+2\pi)i$ on the other edge?
Since we want the logarithm to be continuous, we have to take the interior of the stripe (removing both edges, ${\rm int\,} S_a=\{x+yi\mid y\in (a,a+2\pi)\}$), else by the above, on the border, by continuity we would have
$$x+ai=\log(e^{x+ai})=\log(e^{x+(a+2\pi)i})=x+(a+2\pi)i\,.$$
(For the principal branch, $a=-\pi$ is taken.)
