As @egreg pointed out in the comments, to avoid ambiguity a pair of parentheses is recommended. Or maybe you could try $\max_{x\ge2} -{\log x}$ instead of the full expression $\max\{ \, -{\log x} : x\ge2 \, \}$.
You mentioned you wanted a unary negation, so you have to use $-{\log x}$, and even $-\log x$ is incorrect.
In the $-\log x$ example, \log is of class \mathop, and the minus sign (which is not a binary operation here) will be rendered as an ordinary object (class \mathord - see this answer to learn about math classes). And so a thin space \, is added between them (because a thin space is always inserted between \mathord and \mathop - see this answer - think $\sin x\cos x$ for example).
Therefore, to truly get unary negation, you must make sure whatever you are negating is itself an ordinary object. This can be done by simply enclosing \log x within a pair of {}. Same idea applies to $-{\sin x}$, $\tan x = {\sin x}/{\cos x}$, etc.
\max(-\log x)– egreg Jun 03 '18 at 19:50$\max -\log x$as opposed to$\max -b$. – root Jun 03 '18 at 19:54\max\log x, because there's no binding of the variable. I can suggest\max-{\log x}, – egreg Jun 03 '18 at 19:59\max\{\log x:x>0\}(or whatever interval you compute the maximum over. – egreg Jun 03 '18 at 20:02\max{}-\log x? I dislike, but... – Sigur Jun 03 '18 at 20:04$$\max_{x\in X} -\log x$$. @egreg is that what you mean by binding? – root Jun 03 '18 at 20:08\[ \max-\log x \]\[ \max{}-\log x \]. https://imgur.com/a/1MCHvT7 – Sigur Jun 03 '18 at 20:13