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Question

# Value of x satisfying the equation log(2x+3)(6x2+23x+21)=4−log(3x+7)(4x2+12x+9)

A
13
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B
14
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C
15
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D
16
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Solution

## The correct option is B −14log(2x+3)(6x2+23x+21)=4−log(3x+7)(4x2+12x+9)⇒log(2x+3)(2x+3)(3x+7)=4−log(3x+7)(2x+3)2⇒log(2x+3)(2x+3)+log(2x+3)(3x+7)=4−log(3x+7)(2x+3)2⇒1+log(2x+3)(3x+7)=4−log(3x+7)(2x+3)2Let α=log(2x+3)(3x+7)⇒1α=log(3x+7)(2x+3)2⇒1+α=4−2α⇒α+2α=3⇒α2−3α+2=0⇒(α−1)(α−2)=0⇒α=1 or α=2⇒log(2x+3)(3x+7)=1 or log(2x+3)(3x+7)=2⇒3x+7=2x+3 or 3x+7=(2x+3)2⇒x=−4 or 3x+7=4x2+12x+9⇒x=−4 or 4x2+12x+9−3x−7=0⇒x=−4 or 4x2+9x+2=0⇒x=−4 or (4x+1)(x+2)=0⇒x=−4 or x=−14 or x=−2 When x=−4 then 2x+3=−5<0Hence log(2x+3)(6x2+23x+21) is not defined.When x=−14 then 2x+3=2×−14+3=−12+3=53>0Hence log(2x+3)(6x2+23x+21) is defined.When x=−2 then 2x+3=2×−2+3=−4+3=−1<0Hencelog(2x+3)(6x2+23x+21) is not defined.∴x=−14

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