Question

Consider a logical address with a page size of 8 kb. how many bits must be used to represent the page offset in the logical address?

Answer 1

8 kilobytes have addresses from hex 0x0000 to 0x1fff (=8191). You would need 13 bits for that. 2^13-1 = 8191. If you write the highest address in binary, you get:

0001 1111 1111 1111. Just count the number of 1's.

If you find this difficult to visualise, write down all logical addresses of a small page size of e.g. 8 bytes. You can see you can make all addresses with 3 bits.