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?
 A: $\log_2x-8-9\log_x2=0$
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 $\log_2x=9 \Longrightarrow x=512$.
A: HINT:
Use $$\log_ab=\frac{\log b}{\log a}$$ to form a Quadratic Equation in $\log x$ 
A: From the basic properties, rewrite $$log_2x=8+9log_x2$$ to get $$\frac{\log (x)}{\log (2)}=8+9\frac{\log (2)}{\log (x)}$$ Now, define $y=\log(x)$, so $$\frac{y}{\log (2)}=8+9\frac{\log (2)}{y}$$ Reduce to same denominator, expand, get a quadratic in $y$, solve and go back from the solution of $y$ to the solution for $x$.
I am sure that you can take from here.
A: let, $\log_2 x=y$,then the equation becomes
$$y=8+\frac{9}{y}$$
Follow hints.
Thus, the equation reduces to $$y^2-8y-9=0$$
easy,right?
A: 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}$
