Final answer:
The logical address requires 17 bits, with 6 bits to identify the page number among the 64 pages, and 11 bits to represent the offset within the page of 2,048 bytes.
Step-by-step explanation:
To determine how many bits are required in the logical address for a logical address space of 64 pages with 2,048 bytes per page, mapped onto a physical memory of 16 frames, we need to calculate the bits needed for the page number and the offset within the page.
Firstly, since there are 64 pages, we need to know the number of bits to represent these pages. We can calculate this by finding the log base 2 of 64, which is 6 (because 2^6 = 64). This means we need 6 bits to represent the page number.
Secondly, for the offset within each page, we have 2,048 bytes per page. Again, we find the log base 2 of 2,048 to get the number of bits, which is 11 (because 2^11 = 2,048). Therefore, we need 11 bits for the offset within the page.
In total, the logical address requires 6 bits for the page number plus 11 bits for the offset, which is equal to 17 bits.