Analysis of the algorithms

Question

T(n) = 3T( n/3 ) + n/2

The answer to the above question says that case 2 of masters theorem is applied here. How is it so?

Answer 1

T(n) = aT(n/b) +cn^k

here,  T(n) = 3T( n/3 ) + n/2

k=1 ; c =1/2 ;  a=3 ; b=3

a = b^k

3  = 3¹

Complexity = n^klogn

= O(nlogn)

Answer 2

Case 2: f(n) = n^clog^kn   c=  $\log_b a$

c=1

 $\log_b a$ =1