How to solve this equation?5 $$\log\left(\frac{\ 2}{\sqrt {2-\sqrt 3}}\right) x^2-4x-2=\log\left(\frac{\ 1}{\ (2-\sqrt 3}\right) x^2-4x-3$$
I have been trying to resolve this equation for quite some time now, but for some reason the problem refuses to yield a solution. Some hints will be appreciated.
 A: To solve the problem 
$x^2 \log (\frac{2}{\sqrt {2-\sqrt3} }) - 4x - 2 = x^2 \log (\frac{1}{2-\sqrt3}) - 4x - 3$
, we'll first add 
$4x+2$ 
to both sides, giving 
$x^2 \log (\frac{2}{\sqrt {2-\sqrt3} }) = x^2 \log (\frac{1}{2-\sqrt3}) - 1$
.
We'll then use the property of logarithms which states:
$\log(\frac{a}{b}) = \log(a)-\log(b)$.
We'll apply that to 
$x^2 \log (\frac{2}{\sqrt {2-\sqrt3} })$ 
first, giving 
$\log(2)*x^2-\log(\sqrt {2-\sqrt3} )*x^2$
. Applying the rule of powers inside logarithms, 
$\log(a^b) = b*\log(a)$
, we then obtain 
$\log(2)*x^2-\frac{1}{2}\log(2-\sqrt3)*x^2$
.
Our expression right now should look like this: 
$\log(2)*x^2-\frac{1}{2}\log(2-\sqrt3)*x^2 = x^2\log(\frac{1}{2-\sqrt3}) - 1$
We'll then solve the right side just like the left. We now have 
$\log(2)*x^2-\frac{1}{2}\log(2-\sqrt3)*x^2 = \log(1)*x^2 - \log(2-\sqrt3)*x^2-1$
.
Adding 
$\log(2-\sqrt3)*x^2$ 
to both sides gives
$\log(2)*x^2+\frac{1}{2}\log(2-\sqrt3)*x^2=\log(1)*x^2-1$
.
Let's now factor out $x^2$, giving us
$x^2(\log(2)-\log(1)+\frac{1}{2}\log(2-\sqrt3) = -1$
. 
$\log(1) = 0$, which we know because $b^0 = 1$. We now have $x^2(\log(2)+\frac{1}{2}\log(2-\sqrt3)) = -1$.
We'll now divide by $(\log(2)+\frac{1}{2}\log(2-\sqrt3))$, allowing us to see that $x^2 = -\frac{1}{\log(2)+\frac{1}{2}\log(2-\sqrt3))}$.
Unfortunately, I can't help you any more until I know what base the $\log$ is. $\log_{10}(2)$ is way different from $\log_{e}(2)$, making it very hard to get any other results from this problem.
If you have any questions, which by my lousy and messy proof you probably do, go ahead and ask them!
