Pre-calculus algebra logarithm question

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?

Answer 1

$\log_2x-8-9\log_x2=0$

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 $\log_2x=9 \Longrightarrow x=512$.

Answer 2

HINT:

Use $$\log_ab=\frac{\log b}{\log a}$$ to form a Quadratic Equation in $\log x$

Answer 3

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.

Answer 4

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}$

Answer 5

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?