Question

Solving $lo{g}_x\left(3\right)\cdot lo{g}_{3x}\left(3\right)=lo{g}_{9x}\left(3\right)$ we get $x=m^{\pm \sqrt{n}}$. Find $m+n$ ?

Answer 1

${\log }_{x}3{\log }_{3x}\left(3\right)=lo{g}_{9x}\left(3\right)$
$\frac{1}{{\log }_{3}x}\frac{1}{{\log }_{3}3x}=\frac{1}{{\log }_{3}9x}$
${\log }_{3}9x={\log }_{3}x{\log }_{3}3x$
$\Rightarrow {\log }_{3}9+{\log }_{3}x={\log }_{3}x({\log }_{3}3+{\log }_{3}x)$
$\Rightarrow 2+{\log }_{3}x={\log }_{3}x+({\log }_{3}x{)}^2$
$\Rightarrow ({\log }_{3}x{)}^2=2$
$\Rightarrow x=3^{\sqrt{2}},3^{-\sqrt{2}}$
comparing with $m^{\pm \sqrt{n}}$
$m=3$ and $n=2$
$m+n=5$