what is the minimum value of f(x) where
f(x)=(logax+logbx+logcx) and a>b>c>1..
option
a. 3
b 2
c 1
d 4

• here a,b,c are base for log..they aren't multiplied with x.
  so logax=logex/logea;
  f(x)= (logx/loga + logx/logb+ logx/logc);
  = logx * (1/loga + 1/logb + 1/logc);
  for minimum value of f(x) log values must me max 1
  =1*3
  = 3
• 8 years agoHelpfull: Yes(11) No(2)
• a it can be written as log x(1/log a + 1/log b + 1/ log c) log a ,b,c can be max 1 at that point fx is min so answer is 3
• 8 years agoHelpfull: Yes(3) No(4)
• but here given that c>1 and a>b>c so how can it be possible to take 1/loga=1/logb=1/logc=1
• 8 years agoHelpfull: Yes(2) No(1)
• if we take x=1 then f(x) is becoming 0.
  please rectify if i am wrong.
• 8 years agoHelpfull: Yes(0) No(1)
• And after adding these three values, when we multiply it with log x; then what will be the minimum answer (-:
• 7 years agoHelpfull: Yes(0) No(0)