Sorry, but that's nonsense. If you say that searching for an element in an unsorted list becomes O(log n) if you stop searchng after you've found the element, you actually say that every element in a list is located among it's first log(n) elements, (where n is the length of the list) and everybody knows that the elements of a list reside within the whole list, that's their definition. So your assumption would be true if there existed some "a" and some "n" for which logan = n, but there is no such a.
Btw.: O(n) does not mean that you don't stop searching a list after you've found the thing you're looking for, of course you do that, but you must expect that you have to search the whole list if the element you're looking for is the last element or not in the list.
But you _can_ make list search O(log n) by keeping the list sorted and then doing binary search.
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