As mentioned, the actual number has O(N) bits and thus can't actually be computed in O(log N) time. So if you just want an approximation, you may as well use Binet's formula. [1]
However, one thing this algorithm can do that Binet's formula can't is compute the N-th Fibonacci number modulo some constant M in O(log N) time. So you could very efficiently compute, say, the last 12 digits of F_123456789. A lot of programming contests (TopCoder, Google Code Jam, etc.) will ask for the answer modulo some constant to take advantage of this trick.