Question

For $a>0$ and $a\ne 1,$ the roots of the equation ${\log }_{ax}a+{\log }_x{a}^2+{\log }_{a^{2}x}{a}^3=0$ are

• A. $a^{-4/3}$
• B. $a^{-3/4}$
• C. $a$
• D. $a^{-1/2}$

Answer 1

Answer: (A), (D)

The correct options are
A $a^{-4/3}$
D $a^{-1/2}$

Clearly, $ax>0,x>0,a^{2}x>0$
and $ax\ne 1,x\ne 1,a^{2}x\ne 1$
$\Rightarrow x>0$ and $x\ne \frac{1}{a},1,\frac{1}{a^2}$

Now, ${\log }_{ax}a+{\log }_x{a}^2+{\log }_{a^{2}x}{a}^3=0$
$\Rightarrow \frac{1}{{\log }_{a}ax}+\frac{2}{{\log }_{a}x}+\frac{3}{{\log }_a{a}^{2}x}=0$
$\Rightarrow \frac{1}{1+{\log }_{a}x}+\frac{2}{{\log }_{a}x}+\frac{3}{2+{\log }_{a}x}=0$

Let $y={\log }_{a}x$
Then $\frac{1}{y+1}+\frac{2}{y}+\frac{3}{y+2}=0,y\ne -2,-1,0$

$\Rightarrow \frac{y(y+2)+2(y+1)(y+2)+3y(y+1)}{y(y+1)(y+2)}=0$

$\Rightarrow \frac{6y^2+11y+4}{y(y+1)(y+2)}=0$

$\Rightarrow 6y^2+11y+4=0\Rightarrow (2y+1)(3y+4)=0\Rightarrow y=-\frac{1}{2},-\frac{4}{3}$
$\Rightarrow {\log }_{a}x=-\frac{1}{2},-\frac{4}{3}$
$\therefore x=a^{-1/2},a^{-4/3}$