Question

Solve ${\log }_{a}ax{\log }_{x}ax={\log }_{a^2}\left(\frac{1}{a}\right)$

Answer 1

The roots must satisfy the condition x>0, $x\ne 1$
The equation can be written as
$[lo{g}_{a}a+lo{g}_{a}x][lo{g}_{x}a+lo{g}_{x}x]=lo{g}_{a^2}{a}^{-1}$
or $(1+lo{g}_{a}x)(\frac{1}{lo{g}_{a}x}+1)=-\frac{1}{2}$
Now put $lo{g}_{a}x=y$ Then
$\frac{(1+y{)}^2}{y}=-\frac{1}{2}or2y^2+5y+2=0$
or $(y+2)(2y+1)=0$
This gives $y=-2or-\frac{1}{2}$, that is,
$lo{g}_{a}x=-2or-\frac{1}{2}$
Hence $x=a^{-2}or{a}^{\frac{-1}{2}}$