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Question

If a,b,c are distinct positive numbers,each different from 1, such that
[logbalogca−logaa]+[logablogcb−logbb]+[logaclogbc−logcc]=0, then abc=

A
1
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B
2
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C
3
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D
None of these
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Solution

The correct option is A 1
We have,
[logbalogcalogaa]+[logablogcblogbb]+[logaclogbclogcc]=0

[logbalogca1]+[logablogcb1]+[logaclogbc1]=0

logbalogca+logablogcb+logaclogbc3=0

logbalogca+logablogcb+logaclogbc=3

logalogb×logalogc+logbloga×logblogc+logcloga×logclogb=3

(loga)2logblogc+(logb)2logalogc+(logc)2logalogb=3

(loga)3+(logb)3+(logc)3=3(logalogblogc)
(loga)3+(logb)3+(logc)33(logalogblogc)=0

We know that
(x3+y3+z3)3xyz=(x+y+z)(x2+y2+z2xyyzzx)

Therefore,
(loga+logb+logc)((loga)2+(logb)2+(logc)2logalogblogblogclogcloga)=0

(loga+logb+logc)=0

logabc=0
abc=e0
abc=1

Hence, this is the answer.

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