The value of x which satisfies the equation log4+(1+12x)log3=log(x√3+27) is
A
x∈ϕ
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B
1
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C
4
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D
none of these
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Solution
The correct option is D none of these log4+(1+12x)log3=log(x√3+27) ⇒log4+log3+log312x=log(31x+27) ⇒log(12.312x)=log(31x+27)[∵loga+logb+logc=log(abc)] ⇒12.312x=31x+27 Let 312x=y ⇒y2−12y+27=0⇒y=9 or y=3 ⇒x=14 and x=12