2015/065) What is the value of f(14400) from the following case?

• Function f from the positive integers to the positive integers satisfies the following conditions:
  1. $f(xy)=f(x)+f(y)-1$ for any pair of positive integers x and y.
  2. $f(x)=1$ holds for only finitely many x.
  3. $f(30)=4$
  
  Solution
  
  From (1) we find that
  $f(30)=f(2*15)=f(2)+ f(15)-1$
  = $f(2)+ f(3*5)-1$
  = $f(2) + f(3) + f(5) – 2$
  
  With (3) this yields:
  $f(2)+f(3)+f(5)-2=4$
  $f(2)+2f(3)+f(5)=6 (4)$
  Lemma (1)
  
  
  Neither f(2), nor f(3), nor f(5) can be 1.
  
  Proof
  Suppose  one of them is 1, say $f(2)$, then $f(2^k)=kf(2)-k+1=1$
  That means that infinitely many numbers x have $f(x)=1$
  This is a contradiction with (2).
  Combining lemma with [1] tells us that f(2)=f(3)=f(5)=2.
  Lemma (2)
  
  $f(a^n) = n f(a) – (n-1)$
  
  
  
  the above can be proved by induction
  
  
  
  It follows from (1) that:
  $f(14400)=f(144 * 100)$
  = $f( 2^ 4 * 3^ 2 * 2^2 * 5^2)$
  = $f(2^6 * 3^2 * 5^2)$
  = $6f(2) + 2f(3) + 2f(5)- 9 = 6 * 2 – 5 + 2 * 2 -1 + 2 * 2 – 1 - 2 = 10*2 - 9 = 11$
  I have solved the same at https://in.answers.yahoo.com/question/index?qid=20131011072645AAoFbsS