given a,b,c are in GP and a < b < c.
calculate how many solution exist for this inequality
(log(a) + log(b) + log(c) ) = 6..?

• Since a , b , c are in G.P;
  
  So,we can assume a=(b/r) ; and c = br ; where r is the common ratio.
  
  So, log(a) + log(b) + log(c) = log(b/r) + log(b) + log(br) =
  
  =log(b) - log(r) + log(b) + log(b) + log(r)= 3*log(b) = 6 [Given in the problem]
  
  So, log(b) = 2.
  
  Now,base of the log isn't mentioned.So, for a general case if base is K , then
  
  b= (K^2);
  
  and so,
  
  a=[(K^2)/r];
  
  c=(K^2)*r;
  
  Now, since in the problem,it is mentioned that a < b < c
  So, we have to another criterion for the value of r and that will be
  
  r >1 ..That will do..
  
  Now,since in general practice,we assume 10 as the base of logarithm, so putting
  
  K=10,We get
  
  a=(100/r);
  
  b=100;
  
  c=(100*r);
  
  Where r is greater than 1 ,i.e, r > 1 ..
  
  If any1 of u have any doubt/confusion or can find any loop-hole then never
  
  hesitate to mention it..
  
  Thanks to all.
  
• 11 years agoHelpfull: Yes(29) No(2)
• log(abc)=6
  abc=10^6 because of b^2=ac=100,
  b^3=10^6,
  b=10^2=100
  and the combinations are
  (2,100,5000), (4,100,2500), (5,100,2000), (10,100,1000), (50,100,200),(25,100,400),(20,100,500).
  
• 11 years agoHelpfull: Yes(26) No(2)
• a,b,c are in g.then we know that log(a),log(b),log(c) will in A.P.
  and now check the equation log(a)+log(b)+log(c)=6.then log(a) will be 1,log(b) will be 2,and log(c) will be 3,so in this case a=e^1,b=e^2,c=e^3.
  when will put log(a)=0,log(b)=2,log(c)=4.
  so another answer will be e^0,e^2,e^4.so there are many solutions for this equality.but if we consider a,b,c are as integer then only answers are possible.
• 11 years agoHelpfull: Yes(3) No(11)
• @sdeep bhattacharya..
  wats the answer ??but your solution s very clear,
  when we submit those values we ll get 6 obviously...how many solutions they asked na??is it 100??
• 11 years agoHelpfull: Yes(2) No(0)
• only one solution bcoz condtn of a ,b,c in gp shud also be satisfied.
• 11 years agoHelpfull: Yes(1) No(13)
• @pankaj kumar there are totally 8 solutions...another possible combination is (40,100,250)..
• 9 years agoHelpfull: Yes(1) No(1)
• log(abc)=6
  abc=10^6 because of b^2=ac=100,
  b^3=10^6,
  b=10^2=100
  
  b^2 = ac (as a,b,c are in g.p)
  therefore ac=100*100=10000
  therefore possible combination are
  (1,10000)100
  (2,5000)100
  (4,2500)100
  (5,2000)100
  (8,1250)100
  (10,1000)100
  (16,625)100
  (20,500)100
  (25,400)100
  (40,250)100
  (50,200)100
  (80,125)100
  
  therefore total combinations are 12.
  
  shorcut
  ab=10000
  no.of factors of it =25
  exclude 100 (becoz then a=b=c=100)
  remaining 24 factors
  and we have to make it in set of 2 so possible combination is 24/2 = 12 ans
  
  
• 8 years agoHelpfull: Yes(1) No(0)
• hi SDeep Bhattacharya,check this solution.
• 11 years agoHelpfull: Yes(0) No(15)
• log(abc) = 6,
  abc = 10^6.
  then satisying gp series,
  one solution.
  10,10^2,10^3.
• 10 years agoHelpfull: Yes(0) No(0)
• log(abc)=6
  abc=10^6 because of b^2=ac=100,
  b^3=10^6,
  b=10^2=100
  and the combinations are
  (2,100,5000), (4,100,2500), (5,100,2000), (10,100,1000), (50,100,200),(25,100,400),(20,100,500).
  
  
• 8 years agoHelpfull: Yes(0) No(0)