The answer is given. Please explain in detail why a=99.

The recurrence T(n) = 10T(n/3) + n2 describes the running time of an algorithm A. A competing algorithm A, has a running time of T'(n) = aT'(n/9 + n2. What is the largest integer value for a such that A' is asymptotically faster than A? The time complexity of Algorithm A is T(n) = (nl glo). For any integer a such that a = 99, the time complexity of algorithm A, is T'(n)-6(nloe ) = (nlog,99), which grows slower than T(n). Therefore the answer is a 99 Grading: 3 pts for the relationship log alogs 10; 2 pts for the