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_2 x=8+9log_x 2 Can someone please help me out?

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:
${\mathrm{log}}_{2}x=8+9{\mathrm{log}}_{x}2$
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lufi8c
${\mathrm{log}}_{2}x-8-9{\mathrm{log}}_{x}2=0$
Using ${\mathrm{log}}_{x}2=\frac{1}{{\mathrm{log}}_{2}x}$ and setting $u={\mathrm{log}}_{2}x$ we get:
$u-8-\frac{9}{u}⟺{u}^{2}-8u-9=0⟺\left(u-9\right)\left(u+1\right)=0$
This gives ${\mathrm{log}}_{2}x=-1⟹x=\frac{1}{2}$ or ${\mathrm{log}}_{2}x=9⟹x=512$

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imchasou
HINT:
Use
${\mathrm{log}}_{a}b=\frac{\mathrm{log}b}{\mathrm{log}a}$
to form a Quadratic Equation in $\mathrm{log}x$

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