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Question

# If a,b,c are distinct positive numbers each different from 1 such that [logb a logc a−loga a]+[loga b logc b−logb b]+[loga c logb c−logc c]=0 then abc

A
1
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B
2
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C
3
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D
None
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Solution

## The correct option is A 1Changing all the logarithms to base α(α>0,α≠1) given equation yields ∑[xy.xz−1]=0wherex=logα α etc. or x2yz+y2zx+z2xy=3 or x3+y3+z3−3xyz=0 or [x+y+z][x2+y2+z2−xy−yz−zx]=0 ... (1) Since x≠y≠z, we have x2+y2+z2−xy−yz−zx =12[(x−y)2+(y−z)2+(z−x)2]≠0 Hence we conclude from (1)that\\ (x+y+z)=0, that is logα a+logα b+logαc=0 logα abc = 0 or abc =1

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