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Question

Value of x satisfying the equation
log(2x+3)(6x2+23x+21)=4log(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 14
log(2x+3)(6x2+23x+21)=4log(3x+7)(4x2+12x+9)
log(2x+3)(2x+3)(3x+7)=4log(3x+7)(2x+3)2
log(2x+3)(2x+3)+log(2x+3)(3x+7)=4log(3x+7)(2x+3)2
1+log(2x+3)(3x+7)=4log(3x+7)(2x+3)2
Let α=log(2x+3)(3x+7)1α=log(3x+7)(2x+3)2
1+α=42α
α+2α=3
α23α+2=0
(α1)(α2)=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+93x7=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<0
Hence log(2x+3)(6x2+23x+21) is not defined.
When x=14 then 2x+3=2×14+3=12+3=53>0
Hence log(2x+3)(6x2+23x+21) is defined.
When x=2 then 2x+3=2×2+3=4+3=1<0
Hencelog(2x+3)(6x2+23x+21) is not defined.
x=14


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