Latest Elitmus Aptitude Question SOLUTION: 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..?

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(#M40010588) ELITMUS question NumberTheory Keep an EYE

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..?

Asked In Elitmus sandeep pandey (11 years ago)

Unsolved Read Solution (10)

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SDeep Bhattacharya  (11 years ago)

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.

Is this solution Helpfull? Yes (29) | No (2)  

PANKAJ KUMAR TRIPATHI  (11 years ago)

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).

Is this solution Helpfull? Yes (26) | No (2)  

ankur trivedi  (11 years ago)

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.

Is this solution Helpfull? Yes (3) | No (11)  

deepika  (11 years ago)

@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??

Is this solution Helpfull? Yes (2) | No   

shweta  (11 years ago)

only one solution bcoz condtn of a ,b,c in gp shud also be satisfied.

Is this solution Helpfull? Yes (1) | No (13)  

NAGENDRA P  (9 years ago)

@pankaj kumar there are totally 8 solutions...another possible combination is (40,100,250)..

Is this solution Helpfull? Yes (1) | No (1)  

ravi prakash  (8 years ago)

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

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ankur trivedi  (11 years ago)

hi SDeep Bhattacharya,check this solution.

Is this solution Helpfull? Yes | No (15)  

naga sarath chand  (10 years ago)

log(abc) = 6,
abc = 10^6.
then satisying gp series,
one solution.
10,10^2,10^3.

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raja  (8 years ago)

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).

Is this solution Helpfull? Yes | No   

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