These have more than 2 factorizations, but good idea if you can find how to do it
There is a linear relationship between log n and the number of digits or bits of n. (Using base 2 or e or 10 would just give a different scale ). So a log n algorithm would have solving routine that might:
-narrow down (magnify) search area by 10 times per step, so step 1 tells you first digit, step two tells you second digit, and so on
-narrow down by 2 times per step, e.g a binary tree. If there is a Tree of Knowledge, and the first step is whether e is even or odd, and the following steps iterate some similar decision, then this would be log n
-being able to decide first bit, second bit, third bit in constant time each
One point about what VQC said, either the VQC works in log n time (really we mean log c, since it is c which is given in our problem), and VQC misspoke by saying log n where n is the size of the digits of the input. If VQC meant that literally and wasn't being sloppy then since the number of digits is already log of the number, the VQC's speed is log (log c).