Why is my log in the std namespace?

Question

In the code below, I define a trivial log function. In main I try not to call it; I call std::log. Nevertheless, my own

Answer 1

What happens, I expect, is that std::log simply delegates to ::log. Unfortunately, ::log only provides a float overload, and you kindly provide a double overload, making yours a better match. But I still don't see how it even gets considered in the overload set.