KVPY 2015 9

$\begin{aligned} \log_{3x-1}(x-2) & = \log_{9x^2-6x+1}(2x^2-10x-2) \\ \frac{\log (x-2)}{\log (3x-1)} & = \frac{\log (2x^2-10x-2)}{\log (9x^2-6x+1)} \\ & = \frac{\log (2x^2-10x-2)}{\log (3x-1)^2} \\ & = \frac{\log (2x^2-10x-2)}{2 \log (3x-1)} \\ \Rightarrow 2 \log (x-2) & = \log (2x^2-10x-2) \\ \log (x-2)^2 & = \log (2x^2-10x-2) \\ (x-2)^2 & = 2x^2-10x-2 \\ x^2 - 4x + 4 & = 2x^2-10x-2 \\ \Rightarrow x^2 - 6x - 6 & = 0 \\ \Rightarrow x & = \frac {6 \pm \sqrt{36+24}}{2} \\ & = \boxed{3 + \sqrt{15}} \end{aligned}$

The negative value of $x$ is rejected because $\log_{3x-1}(x-2)$ and $\log_{9x^2-6x+1}(2x^2-10x-2)$ are not defined.

Great problem!

We have, $\begin{aligned} \quad \log_{(3x - 1)} (x - 2) = \log_{(9x^2 - 6x + 1)} (2x^2 - 10x - 2) \\ \implies \log_{(3x-1)}{(x-2)}=\log_{(3x-1)^{2}}{2x^2-10x-2} \\ \implies \log_{(3x-1)}{(x-2)}=\log_{(3x-1)}{\sqrt{(2x^2-10x-2)}} \\ \implies x-2=\sqrt{2x^2-10x-2} \\ \implies x^2-4x+4=2x^2-10x-2 \\ \implies x^2-6x-6=0 \end{aligned}$ We can now use the quadratic formula. $\implies x=\dfrac{6 \pm \sqrt{36+24}}{2} \implies x=\dfrac{6 \pm \sqrt{60}}{2} \implies x=3 \pm \sqrt{15}$ But $3-\sqrt{15}$ is negative and $\log_{3x-1}{x-2}$ is undefined for negative $x$ . $\boxed{\therefore x=3+\sqrt{15}}$