How many 32-bit integers can be stored in a 16-byte cache line?.

4

Step-by-step explanation:

1 byte = 8 bits
=> 16 byte = 16 x 8 = 128 bits
So it can hold 128 : 32 = 4 integers.

logv3(x+2)=1

rewrite in exponential form:

logb(x) = y is equal to b^y =x

b = 3, y = 1

now solve for x:

3^1 = x+2)

x = 3-2

x = 1

see the picture for the graph.