Question

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

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

Answer 1

Answer: (B)

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

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

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

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

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

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

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

Put ${\log }_{a}x=t$, then,

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

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

$2+t+4+2t+3+3t=0$

$6t+9=0$

$t=-\frac{3}{2}$

${\log }_{a}x=-\frac{3}{2}$

$x=a^{-\frac{3}{2}}$