Question

$lo{g}_{x}a+-lo{g}_{ax}a+3lo{g}_{a^{2}x}a=0$

Answer 1

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

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

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

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

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

Let $y={\log }_{a}x$

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

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

$\Rightarrow (1+y)(2+y)-y(2+y)+3y(1+y)=0$

$\Rightarrow (2+y+2y+y^2)-(2y+y^2)+(3y+3y^2)=0$

$\Rightarrow 2+3y+y^2-2y-y^2+3y+3y^2=0$

$\Rightarrow 3y^2+4y+2=0$

Now by quadratic formula, we have

$y=\frac{-4\pm \sqrt{16-24}}{6}$

$\Rightarrow y=\frac{-2\pm 2\sqrt{2}i}{3}$

Therefore,

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

$\Rightarrow x=a^{\frac{-2\pm 2\sqrt{2}i}{3}}$

Hence $x=a^{\frac{-2\pm 2\sqrt{2}i}{3}}$.