Range of a log function

Question

Find the range of $f(x)=\ln(\frac{\sqrt{8-x^2}}{x-2})$

My attempt :- $$f(x)=\ln(\frac{\sqrt{8-x^2}}{x-2})$$ $$=0.5\ln(8-x^2)-\ln(x-2)$$
After this I can't solve and not able to find the range.

Answer 1

Hint:

Let $\dfrac{8-x^2}{(x-2)^2}=y$

$\iff x^2(y+1)-4x+4y-8=0$

$x=\dfrac{2\pm\sqrt{52-(4y-2)^2}}2=2\pm\sqrt{13-(2y-1)^2}$

Now as $2<x<2\sqrt2$ and $2-\sqrt{13-(2y-1)^2}\le2$

$x=2+\sqrt{13-(2y-1)^2}$

$\iff2<2+\sqrt{13-(2y-1)^2}<2\sqrt2$

$2<2+\sqrt{13-(2y-1)^2}\implies13-(2y-1)^2\ne0\iff(2y-1)^2<13$

$2+\sqrt{13-(2y-1)^2}<2\sqrt2\implies13-(2y-1)^2<(2\sqrt2-2)^2$