Solve the equation: $\;$ $\dfrac{1}{6} \log_2 \left(x - 2\right) - \dfrac{1}{3} = \log_{\frac{1}{8}} \sqrt{3x-5}$
Given equation: $\;\;$ $\dfrac{1}{6} \log_2 \left(x - 2\right) - \dfrac{1}{3} = \log_{\frac{1}{8}} \sqrt{3x-5}$
i.e. $\;$ $\dfrac{1}{6} \log_2 \left(x - 2\right) - \log_{\frac{1}{8}} \sqrt{3x-5} = \dfrac{1}{3}$
i.e. $\;$ $\dfrac{1}{6} \log_2 \left(x - 2\right) - \dfrac{\log_2 \sqrt{3x-5}}{\log_2 \dfrac{1}{8}} = \dfrac{1}{3}$
i.e. $\;$ $\dfrac{1}{6} \log_2 \left(x - 2\right) - \dfrac{\log_2 \left(3x - 5\right)^{\frac{1}{2}}}{\log_2 2^{-3}} = \dfrac{1}{3}$
i.e. $\;$ $\dfrac{1}{6} \log_2 \left(x - 2\right) - \dfrac{\dfrac{1}{2} \log_2 \left(3x - 5\right)}{-3 \log_2 2} = \dfrac{1}{3}$
i.e. $\;$ $\dfrac{1}{6} \log_2 \left(x-2\right) + \dfrac{1}{6} \log_2 \left(3x-5\right) = \dfrac{1}{3}$
i.e. $\;$ $\dfrac{1}{2} \left[\log_2 \left(x-2\right) + \log_2 \left(3x-5\right)\right] = 1$
i.e. $\;$ $\log_2 \left[\left(x-2\right) \left(3x-5\right)\right] = 2$
i.e. $\;$ $\left(x-2\right) \left(3x-5\right) = 2^2 = 4$
i.e. $\;$ $3x^2 - 11x + 10 = 4$
i.e. $\;$ $3x^2 - 11x + 6 = 0$
i.e. $\;$ $\left(3x-2\right) \left(x-3\right) = 0$
i.e. $\;$ $x = \dfrac{2}{3}$ $\;$ or $\;$ $x = 3$
When $\;$ $x = \dfrac{2}{3}$, $\;$ the term $\;$ $\log_2 \left(x-2\right)$ $\;$ in the given problem becomes
$\log_2 \left(\dfrac{2}{3} - 2\right) = \log_2 \left(\dfrac{-4}{3}\right)$
But, logarithim of a negative number is not defined.
$\therefore \;$ $x = \dfrac{2}{3}$ $\;$ is not a valid solution to the given problem.
When $\;$ $x = 3$, $\;$ the given problem becomes
$\dfrac{1}{6} \log_2 \left(3-2\right) - \dfrac{1}{3} = \log_{\frac{1}{8}} \sqrt{9-5}$
i.e. $\;$ $\dfrac{1}{6} \log_2 1 - \log_{\frac{1}{8}} 2 = \dfrac{1}{3}$
i.e. $\;$ $0 + \log_{\frac{1}{8}} 2 = \dfrac{-1}{3}$
i.e. $\;$ $2 = \left(\dfrac{1}{8}\right)^{\frac{-1}{3}}$
i.e. $\;$ $2 = 8^{\frac{1}{3}}$
i.e. $\;$ $2 = 2$ $\;$ which is true.
$\therefore \;$ The solution to the given equation is $\;\;$ $x = \left\{3 \right\}$