Pre-calculus algebra logarithm question I don't understand how to solve this equation. Been struggling with it and don't know how to start:
$$\log_2x=8+9\log_x2$$
Can someone please help me out?
$\log_2x-8-9\log_x2=0$

Using $\log_x2=\frac{1}{\log_2 x}$ and setting $u=\log_2 x$ we get:

$u-8-\frac{9}{u} \Longleftrightarrow u^2-8u-9=0 \Longleftrightarrow (u-9)(u+1)=0$

This gives $\log_2x=-1 \Longrightarrow x=\frac{1}{2}$ or $\log_2x=9 \Longrightarrow x=512$.
Thanks guys for the hints, however I found another way to solve it and thought I would share it here:
Let $z=log_2x$ and $y=log_x2$
thus we have $x=2^{z}$ and $x=2^{y}$ therefore $2^z=2^{y} , z=y$

plugging that into our equation we get:

$z=8+9y , z=8+9z, z=-1, y=-1$ thus $ x=\frac{1}{2}$