Question

$lo{g}_{2x+3}(6x^2+23x+21)=4-lo{g}_{3x+7}(4x^2+12x+9)$, then $x$ equals

• A. $-4$
• B. $-2$
• C. $-\frac{1}{4}$
• D. None of these

Answer 1

Answer: (C)

If $lo{g}_{a}x=t$ then $x=a^t$ such that $a>0$ and $a\ne 1$
Now, $lo{g}_{2x+3}(2x+3)(3x+7)$
$=4-lo{g}_{3x+7}(2x+3{)}^2\dots (A)$
Now, $2x+3>0$ and $2x+3\ne 1$
$3x+7>0$ and $3x+7\ne 1$
$\therefore 1+lo{g}_{2x+3}(3x+7)=4-lo{g}_{3x+7}(2x+3{)}^2$
$\because lo{g}_a(ab)=1+lo{g}_{a}b$
$\Rightarrow lo{g}_{2x+3}(3x+7)+2lo{g}_{3x+7}(2x+3)-3=0$
Substitute $t=lo{g}_{2x+3}(3x+7)$
$\Rightarrow t+\frac{2}{t}-3=0$
$\Rightarrow t^2-3t+2=0$
$\Rightarrow t=1,t=2$
$\Rightarrow$ 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).