If x is a positive real number different from 1 such that logax,logbxlogcx are in A.P., then
A
b=a+b2
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B
b=√ac
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C
c2=(ac)logab
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D
None of (a),(b),(c) are correct
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Solution
The correct option is Cc2=(ac)logab logax,logbxlogcx are in A.P ⇒2logbx=logax+logcx⇒2logxb=1logxa+1logxc⇒2logxb=logxaclogxalogxc⇒2logxc=logxblogxaclogxa⇒c2=(ac)logab