a,b,c are in increasing order of gp. given (loga+logb+logc)/log6=6 . find the minimum value of c-b?

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• 8 years agoHelpfull: Yes(25) No(13)
• (loga+logb+logc)=6*log6....
  log(a*b*c)=log(6^6)....
  a*b*c=6^6....
  a,b,c are in G.P....G.P. series look like a,ar,ar^2,ar^3.....
  so if we take a=6 and r=6.....
  then a*b*c=(6)*(6*6)*(6*6^2).....look like a*b*c=6*6^2*6^3=6^6...
  a=6,b=36,c=216...
  so c-b=216-36=180....
  
• 8 years agoHelpfull: Yes(11) No(1)
• a,b,c are in gp.
  so b^2=ac
  2logb=loga+logc
  given,
  (loga+logb+logc)/log6=6
  put loga+logc=2logb
  3logb/log6=6
  hence b=36
  for min value of c-b,
  gp=27,36,48
  hence c-b=12
• 8 years agoHelpfull: Yes(8) No(5)
• log(a.ar.ar^2)=6log(6) --> ar=36
  a.r=3.3.2.2, choose a=3.3.2, r=2
  c-b=ar(r-1), 36(r-1), r can minimum be 2.
  Hence min(c-b)=36
• 8 years agoHelpfull: Yes(5) No(0)
• @shabari CHUTIYE agar log(abc)/log(6)=6
  to abc=6^6 hoga
  if u dont know how to solve dont post wrong sol and confuse others
• 8 years agoHelpfull: Yes(4) No(0)
• abc=6^6
  a,b,c are in gp..
  ao b^2=ac
  (6^2)^2=a^2*c^2
  so b=6,a=36,c=36
  now c-b=36-6=30
• 8 years agoHelpfull: Yes(3) No(7)
• log(abc)/log(6)=6 which means that abc =6^6 and by that a=6, b=6^2 and c=6^3.
  taking the difference of 216-36 =180.
  Don't know whether its minimum or not but this is correct.
  Some people posting answers here don't even know what gp is :D
• 8 years agoHelpfull: Yes(3) No(0)
• As we know log(abc)=loga+logb+logc
  given log(abc)/log6=6
  so abc=36
  factors of 36=2,3,6
  for min value we consider c=3,b=2
  therefore ans is 1
• 8 years agoHelpfull: Yes(2) No(10)