How can solve that log

Question

How can solve that logarithms

$\log _{\frac{4}{x}}\left(x^2-6\right)=2$

It's look diffucult to solve

I was solve but stop with

$x^4−6x^2−16=0$

what is next?

Answer 1

Hint: $$(\frac{4}{x})^2=(x^2-6)$$ $$16=x^4-6x^2$$ $$x^4-6x^2-16=0$$ then solve it by using the quadratic formula $$x^2=3\pm\sqrt{9+16}$$ $$x^2=3\pm5$$ $$x^2=8$$ $$x=\pm2\sqrt{2}$$ or $$x^2=-2$$ $$x=\pm\sqrt{2}i$$

Answer 2

Hint: raise $4/x$ to both sides

\begin{align} \log_{\frac{4}{x}}\left(x^2-6\right)&=2\\ \left(\frac4x\right)^{\left(\log_{\frac{4}{x}}\left(x^2-6\right)\right)}&=\left(\frac4x\right)^2\\ \end{align}

Answer 3

it is equivalent to $$\frac{\ln(x^2-6)}{\ln(\frac{4}{x})}=2$$ and further we get $$\ln(x^2-6)=\ln((\frac{4}{x})^2)$$ and this is equivalent to $$x^4-6x^2-16=0$$ Now set $x^2=t$ and solve the quadratic equation.