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 If 1,log331x+2,log34.3x1 are in A.P then, x equals: 

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a
log34
b
1−log34
c
1−log43
d
loge4e

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detailed solution

Correct option is B

First of all logarithmic terms should have been defined, log3⁡31−x+2, so 31−x+2>0 which is true ∀xlog3⁡4.3x−1, so log3⁡4.3x−1 thus 3x>14 Now, terms are in A.P so, 2log3⁡31−x+2=1+log3⁡4⋅3x−1=log3⁡3+log3⁡4.3x−1=log3⁡34.3x−131−x+22=34.3x−131−x+2=34.3x−1=12.3x−31313x+2=12⋅3x−3 Let t=3x3t+2=12t−3, so 3+2t=12t2−3t12t2−5t+3=012t2−9t+4t−3=03t(4t−3)+1(4t−3)=0


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