Answer:
Algorithm 1 uses:
n*log(n) operations.
While algorithm 2 uses:
n^(3/2) operations.
We want to see, as n grows, which algorithm uses fewer operations.
So we would want to first solve:
n*log(n) = n^(3/2)
This will give us the exact value of n such that the number of operations is the same in both algorithms.
dividing both sides by n we get:
log(n) = n^(3/2)/n = n^(3/2 - 1) = n^(1/2)
where we can use:
log(n) = ln(n)/ln(10)
ln(n) = ln(10)*n^(1/2)
This equation actually has no solutions.
This happens because the right side is always larger than the left side.
Then, the same thing happens for our two initial equations:
n^(3/2) is always larger than n*log(n), as you can see in the graph below, where n^(3/2) is represented with the orange graph:
So we can conclude that the fist algorithm uses less operations as n grows.
