Question

For $a>0,a\ne 1$ the number of real values of $x$ satisfying the equation $2{\log }_{x}a+{\log }_{ax}a+3{\log }_{a^{2}x}a=0$ is/are

Answer 1

$2\times \frac{1}{lo{g}_{a}x}+\frac{1}{lo{g}_{a}x}+3\frac{1}{lo{g}_a{a}^{2}x}$

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

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

$\frac{2}{lo{g}_{a}x}+\frac{1}{1+lo{g}_{a}x}+\frac{3}{2\times 1+lo{g}_{a}x}=0$

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

Let $y=lo{g}_{a}x$

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

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

$2(1+y)(2+y)+y(2+y)+3y(1+y)=0$

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

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

Solving the quadratic equation

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

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

$lo{g}_{a}x=\frac{-1}{2}$ and $lo{g}_{a}x=\frac{-4}{3}$

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

So $x$ has two solutions

Therefore Answer is 2.