the fun. Is defined for all positive values for x and y as f(xy)=f(x)+f(y).also f(2)=2,f(3)=3.then f(32/27)=?

• Ans is 1
  as f(32/27)=f((2/3)^3*4)=f((2/3)^3)+f(4)
  f(4)=f(2*2)=f(2)+f(2)=2+2=4
  f((2/3)^3)=f((2/3)^2)+f(2/3)
  f((2/3)^2)=f(2/3)+f(2/3)
  f(2)=f(2/3*3)=f(2/3)+f(3)=>f(2/3)=2-3=-1
  so using above value we get
  f((2/3)^3)=-3
  f(32/27)=-3+4=1
• 12 years agoHelpfull: Yes(27) No(1)
• simply its log property so we can use log property here..
  f(32/27)=f(2^5 / 3^3)=f(2^5 * 3^-3)
  f(2^5)+f(3^-3)=using log,
  5f(2)+(-3)f(3) = 10-9 =1
• 12 years agoHelpfull: Yes(6) No(0)
• ans is always 1 either by log method or by akhil...
• 12 years agoHelpfull: Yes(2) No(2)
• @Pallavi-U r absolutely wrong...use ur brain...
• 12 years agoHelpfull: Yes(2) No(0)
• f(32/27)= f(2^5/3^3)
  f(4)=f(2*2)=4, and similarly forming an A.P
  f(32)=10(2,4,6,8,10)
  f(9)=f(3*3)=3+3=6,
  and hence f(27)=9, (3,6,9)
  f(32*(1/27))= f(32)+f(1/27)
  =10-9=1
  simple...
  
• 12 years agoHelpfull: Yes(2) No(1)
• @ pallavi...wat u hve done..firstly u have 2 study log..
• 12 years agoHelpfull: Yes(1) No(2)
• myn is coming 32.03
• 12 years agoHelpfull: Yes(0) No(10)
• this is a lograthim question. .
  match condition you will automaticly get the function as lograthim.
  now log(32/27) = log32- log 27
  log3+ log2 - log2 - log7
  ans will be 3- log7 :)
  in my consideraion data is wrong. . n it must be as log(32/23)
  then ans will be 0 :)
• 12 years agoHelpfull: Yes(0) No(29)
• ans is 5
• 12 years agoHelpfull: Yes(0) No(4)