$$\log\left(\frac{\ 2}{\sqrt {2-\sqrt 3}}\right) x^2-4x-2=\log\left(\frac{\ 1}{\ (2-\sqrt 3}\right) x^2-4x-3$$

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.

  • $\begingroup$ Please check the denominators of your fractions, I'm not sure which quantities should be contained in the radicals. Other than that, it's a quadratic (and the linear terms vanish). $\endgroup$ – Michael Burr Dec 28 '15 at 21:31
  • $\begingroup$ In the original post, were the bases varying? And if so, what were they? $\endgroup$ – André Nicolas Dec 28 '15 at 21:33
  • $\begingroup$ Where you wrote $\sqrt(2-\sqrt3$, did you mean $\sqrt{2-\sqrt3}$ or $\sqrt2-\sqrt3$ or something else? ${}\qquad{}$ $\endgroup$ – Michael Hardy Dec 28 '15 at 22:44
  • $\begingroup$ please rephrase the question @superman $\endgroup$ – Nebo Alex Dec 29 '15 at 11:04

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!


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