Suppose that you have two different algorithms for solving a problem. to solve a problem of size n, the first algorithm uses exactly n(log(n)) operations and the second algorithm uses exactly n3=2 operations. as n grows, which algorithm uses fewer operations? justify your answer using definitional proof.

So we have two expressions n(log(n)) and 3n = 2. The second expression involves less operation. From the second expression we have n = 2/3. If the RHS and LHS will be balanced, then the value of n cannot exceed 2/3. Sowe have no operation at that point. On the other hand when n = 10 we have 10(log(10)) = 1. When n is increased to 100. We have 100(log(100)) = 200 and when n = 300 we have 300(log(300)) = 300 *2.47 = 743.1 As n increase, the logarithmic function also increases, there's no end in sight; on the other hand as n increase in the first operation, once it reaches 2/3, it stops. So the second has fewer operations.