Question

For $x,a>0$ the root(s) of the equation ${\log }_{ax}a+{\log }_x{a}^2+{\log }_{a^{2}x}{a}^3=0$ is (are) given by

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

Answer 1

Answer: (A), (C)

$a^{\frac{-1}{2}}$
$a^{\frac{-4}{3}}$

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

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

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

$\frac{1}{{\log }_{a}a+{\log }_{a}x}+\frac{2}{{\log }_{a}x}+\frac{3}{{\log }_a{a}^2+{\log }_{a}x}=0$

$\frac{1}{1+{\log }_{a}x}+\frac{2}{{\log }_{a}x}+\frac{3}{2{\log }_{a}a+{\log }_{a}x}=0$

$\frac{1}{1+{\log }_{a}x}+\frac{2}{{\log }_{a}x}+\frac{3}{2\left(1\right)+{\log }_{a}x}=0$

$\frac{1}{1+{\log }_{a}x}+\frac{2}{{\log }_{a}x}+\frac{3}{2+{\log }_{a}x}=0$

Put ${\log }_{a}x=y$,

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

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

$2y+y^2+2\left(y^2+3y+2\right)+3y+3y^2=0$

$2y+y^2+2y^2+6y+4+3y+3y^2=0$

$6y^2+11y+4=0$

$y=\frac{-11\pm \sqrt{{\left(11\right)}^2-4\left(6\right)\left(4\right)}}{2\left(6\right)}$

$y=\frac{-11\pm 5}{12}$

$y=\frac{-6}{12},\frac{-16}{12}$

$y=-\frac{1}{2},-\frac{4}{3}$

${\log }_{a}x=-\frac{1}{2}$

$x=a^{-1/2}$

And,

${\log }_{a}x=-\frac{4}{3}$

$x=a^{-4/3}$

Therefore, the roots are $x=a^{-1/2}$ and $x=a^{-4/3}$.