Question

Find the value of $x$, if ${\log }_{(2x+3)}({6x}^2+23x+21)=4-{\log }_{(3x+7)}({4x}^2+12x+9)$

• A. $x=\frac{-1}{2}$
• B. $x=\frac{-1}{4}$
• C. $x=\frac{1}{4}$
• D. $x=\frac{1}{2}$

Answer 1

Answer: (B)

$x=\frac{-1}{4}$

${\log }_{(2x+3)}({6x}^2+23x+21)=4-{\log }_{(3x+7)}({4x}^2+12x+9)\Rightarrow {\log }_{(2x+3)}\left[(2x+3)(3x+7)\right]=4-{\log }_{(3x+7)}{(2x+3)}^2\Rightarrow 1+{\log }_{(2x+3)}(3x+7)=4-2{\log }_{(3x+7)}(2x+3)[\because \log (ab)=\log a+\log b\text{\&}{\log }_{a}a=1]$
Put ${\log }_{(2x+3)}(3x+7)=y$
$\therefore y+\frac{2}{y}-3=0\Rightarrow y^2-3y+2=0\Rightarrow (y-1)(y-2)=0$
$\Rightarrow y=1$ or $y=2$
$\Rightarrow {\log }_{(2x+3)}(3x+7)=1$ or ${\log }_{(2x+3)}(3x+7)=2$
$\Rightarrow 3x+7=2x+3$ or $(3x+7)={(2x+3)}^2$
$\Rightarrow x=-4$ or $3x+7={4x}^2+12x+9$
$\Rightarrow x=-4$ or ${4x}^2+9x+2=0$
$\Rightarrow x=-4$ or $(4x+1)(x+2)=0$
$\therefore x=-2,-4,\frac{-1}{4}$ ...(1)
But, log exists only when ${6x}^2+23x+21>0,{4x}^2+12x+9>0$
$2x+3>0$ and $3x+7>0$
$\Rightarrow x>-\frac{3}{2}$ ...(2)
From (1) and (2)
$\therefore x=-\frac{1}{4}$