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Q.
$log_{2x+3}\, (6x^{2}+23x+21)=4-log_{3x+7} (4x^{2}+12x+9)$, then $x$ equals
Complex Numbers and Quadratic Equations
Solution:
If $log_{a}\,x=t$ then $x=a^{t}$ such that $a>\,0$ and $a \ne1$
Now, $log_{2x+3} (2x+3) (3x+7)$
$=4-log_{3x+7} (2x+3)^{2}\, \dots(A)$
Now, $2x+3>\,0$ and $2x+3 \ne 1$
$3x+7>\,0$ and $3x+7 \ne 1$
$\therefore 1+log_{2x+3} (3x+7)=4-log_{3x+7} (2x+3)^{2}$
$\because log_{a} (ab) =1 +log_{a} b$
$\Rightarrow log_{2x+3} (3x+7)+2 \,log_{3x+7} (2x+3)-3=0$
Substitute $t=log_{2x+3} (3x+7)$
$\Rightarrow t+\frac{2}{t}-3=0$
$\Rightarrow t^{2}-3t+2=0$
$\Rightarrow t=1, t=2$
$\Rightarrow $ Either $2x+3=3x+7$ or $(2x+3)^{2}=3x+7$
$\Rightarrow x=-4$ or $4x^{2}+9x+2=0$
or $x=-2, -\frac{1}{4}$
$\Rightarrow x=-4, -2, -\frac{1}{4}$
From these three values, only $x=-1/4$ satisfies the equation (A).