Multi-level page tables - hierarchical paging

Question

Example question from a past operating system final, how do I calculate this kind of question?

A computer has a 64-bit virtual address space and 2048-byte pages. A page

Answer 1

Since page table must fit in a page, page table size is 2048 bytes and each entry is 4 bytes thus a table holds 2048/4=512 entries. To address 512 entries it requires log2(512)=9 bits. The total number of bits available to encode the entry for each page level is 64-log2(2048)=53 bits (the number of bits of address space minus the page offset bits). Thus the total number of levels required is 53/9=6 (rounded up).

The x86-64 default page table size is 4096 bytes, each page table must fit in a page and a page table entry is 8 bytes. Current CPUs only implement 48 bits of virtual address space. How many page table levels are required?