Question

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

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

Answer 1

Answer: (A), (D)

$a^{-4/3}$
$a^{-1/2}$

${\log }_{ax}a+{\log }_x{a}^2+{\log }_{a^{2}x}{a}^3=0$

$\Rightarrow \frac{\log a}{\log a+\log x}+\frac{2\log a}{\log x}+\frac{3\log a}{2\log a+\log x}=0[\because {\log }_{b}a=\frac{\log a}{\log b}\text{\&}\log (ab)=\log a+\log b]$

$\Rightarrow \log a\{\log x(2\log a+\log x)+2(2\log a+\log x)(\log a+\log x)+3\log x(\log a+\log x)\}=0$

$\Rightarrow \log x(2\log a+\log x)+2(2\log a+\log x)(\log a+\log x)+3d\log x(\log a+\log x)=0$

$\Rightarrow 6{\left(\log x\right)}^2+11\left(\log a\right)\left(\log x\right)+4{\left(\log a\right)}^2=0$

$\Rightarrow 6{\left(\log x\right)}^2+8\left(\log a\right)\left(\log x\right)+3\left(\log a\right)\left(\log x\right)+4{\left(\log a\right)}^2=0$

$\Rightarrow \log x=-\frac{4\log a}{3},-\frac{1\log a}{2}$

$\Rightarrow \log x=\log a^{-\frac{4}{3}},\log a^{-\frac{1}{2}}$

$\Rightarrow x=a^{-\frac{4}{3}},a^{-\frac{1}{2}}$
Ans: A,D