Is $\log(x)$ uniformly continuous on $(\frac{1}{2},\infty)$? I think the answer will be true.
Here's an attempt:
Let us divide the interval in $(\frac{1}{2},1)$ and $[1,\infty)$.
In the domain $[1,\infty)$ the derivatives of $log(x)$ is bounded so it is uniformly continuous.
In the domain $(\frac{1}{2},1)$ the function is continuous and also $lim_{x \to 1^{-}} log(x)$ exists finitely and $lim_{x \to \frac{1}{2}^{+}} log(x)$ also exists finitely.
Hence the function is uc on the entire domain.

Yes, and you don't even need to split. On your interval you have $|\log'x|≤2$.  Hence, by the Mean Value Theorem,
$$
|\log x-\log y|=|\log'\xi|\,|x-y|≤2|x-y|,\qquad\qquad x, y≥\frac12.
$$
