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Question

For a>0,1, the roots of the equation logaxa+logxa2+loga2xa3=0 are given by

A
a4/3
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B
a3/4
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C
a
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D
a1/2
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Solution

The correct options are
B a4/3
D a1/2
logaxa+logxa2+loga2xa3=0

logaloga+logx+2logalogx+3loga2loga+logx=0[logba=logalogb & log(ab)=loga+logb]

loga{logx(2loga+logx)+2(2loga+logx)(loga+logx)+3logx(loga+logx)}=0

logx(2loga+logx)+2(2loga+logx)(loga+logx)+3dlogx(loga+logx)=0

6(logx)2+11(loga)(logx)+4(loga)2=0

6(logx)2+8(loga)(logx)+3(loga)(logx)+4(loga)2=0

logx=4loga3,1loga2

logx=loga43,loga12

x=a43,a12
Ans: A,D

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