Solving T(n) = 3T(n/3)+n/2 using master method I thought I understood the Master Method quite well till I saw this question

$T(n) = 3T(\frac{n}{3})+\frac{n}{2}$ 

My approach:
  $a = 3 ; b=3$ and $f(n) = \frac{n}{2}$

$n^{\log_b{a}}$ = $n^{log_3{3}} = n$

This looked like a classic solution of master method using case 1. 

But since case 1 implies $f(n) = n^{\log_3{3} - ε}$ and I cannot find any way that ε can be represented, I understand that case 1 is not the way. 

However the solution says that this is solvable by case 2 of the master method and the solution is  $T(n) = \theta(n^{log_3{3}} \log^{ n}) $

Case 2 of the master method states that if $f(n) = n^{log_b{a}}  \log^{k}{ n}$ then $T(n) = \theta(n^{log_b{a}} \log^{k+1}{ n}) $

Can someone explain how this is solved using case 2 of the master method and why this fits under case 2?
While applying master's theorem all you have to do is compare which of $f(n)$ or $(n^{\log_b{a}})$ is asymptotically greater. The greater of the two functions determines the solution to the recurrence. There are only possible outcomes for this comparison: 
[1] $(n^{\log_b{a}})$ is asymptotically greater than $f(n)$ 
[2] Both are asymptotically equal 
[3] $f(n)$ is asymptotically greater than $(n^{\log_b{a}})$ 

Yours falls in case 2.