Solving ${\log_{\frac{2}{\sqrt{2-\sqrt{3}}}}(x^2-4x-2)=\log_{\frac{1}{2-\sqrt{3}}}(x^2-4x-3)}$ This was given to me by my Math Teacher almost a year ago and I've not been able to make much progress on it. I am hoping to see it resolved by our community members. $$\large{\log_{\frac{2}{\sqrt{2-\sqrt{3}}}}(x^2-4x-2)=\log_{\frac{1}{2-\sqrt{3}}}(x^2-4x-3)}.$$

Hint
If $x^2-4x-2<0$ or $x^2-4x-3<0$ then the corresponding $x$ cannot be a solution. But forget about that for a second and explain the meaning of the equation below:
$$\log_{\frac{2}{\sqrt{2-\sqrt{3}}}}(x^2-4x-2)=\log_{\frac{1}{2-\sqrt{3}}}(x^2-4x-3).$$
If for an $x$ the equation holds then there must exist some $A$ for which
$$\log_{\frac{2}{\sqrt{2-\sqrt{3}}}}(x^2-4x-2)=A$$
and
$$\log_{\frac{1}{2-\sqrt{3}}}(x^2-4x-3)=A.$$
By the definition of the logarithm one can say that for such an $A$:
$$\left(\frac{2}{\sqrt{2-\sqrt{3}}}\right)^A=x^2-4x-2$$
and
$$\left(\frac{1}{2-\sqrt{3}}\right)^A=x^2-4x-3.$$
Or
$$x^2-4x-\left(2+\left(\frac{2}{\sqrt{2-\sqrt{3}}}\right)^A\right)=0$$
and
$$x^2-4x-\left(3+\left(\frac{1}{2-\sqrt{3}}\right)^A\right)=0.$$
These are two quadratic equations which can be solved for $x$. The solutions will  depend on the parameter $A$. Since the corresponding solutions  will have to be equal one will have equations for the possible $A$'s.
Given a suitable $A$ an $x$ can be calculated. (Finally test the negativity of the argument of the logarithm.) 
