Solve the equation: $\;$ $\log_2 \left(\dfrac{x}{4}\right) = \dfrac{15}{\log_2 \left(\dfrac{x}{8}\right) - 1}$
Given equation: $\;\;$ $\log_2 \left(\dfrac{x}{4}\right) = \dfrac{15}{\log_2 \left(\dfrac{x}{8}\right) - 1}$
i.e. $\;$ $\log_2 x - \log_2 4 = \dfrac{15}{\log_2 x - \log_2 8 - 1}$ $\;\;\;$ $\left[\because \;\; \log_a \left(\dfrac{m}{n}\right) = \log_a m - \log_a n\right]$
i.e. $\;$ $\log_2 x - \log_2 2^2 = \dfrac{15}{\log_2 x - \log_2 2^3 - 1}$
i.e. $\;$ $\log_2 x - 2 \log_2 2 = \dfrac{15}{\log_2 x - 3 \log_2 2 - 1}$
i.e. $\;$ $\log_2 x - 2 = \dfrac{15}{\log_2 x - 3 - 1}$
i.e. $\;$ $\log_2 x - 2 = \dfrac{15}{\log_2 x - 4}$
i.e. $\;$ $\left(\log_2 x\right)^2 - 2 \log_2 x - 4 \log_2 x + 8 = 15$
i.e. $\;$ $\left(\log_2 x\right)^2 - 6 \log_2 x - 7 = 0$
i.e. $\;$ $\left(\log_2 x\right)^2 - 7 \log_2 x + \log_2 x - 7 = 0$
i.e. $\;$ $\log_2 x \left(\log_2 x - 7\right) + 1 \left(\log_2 x - 7\right) = 0$
i.e. $\;$ $\left(\log_2 x - 7\right) \left(\log_2 x + 1\right) = 0$
i.e. $\;$ $\log_2 x = 7$ $\;$ or $\;$ $\log_2 x = -1$
i.e. $\;$ $x = 2^7 = 128$ $\;$ or $\;$ $x = 2^{-1} = \dfrac{1}{2}$
$\therefore \;$ The solution to the given equation is $\;\;$ $x = \left\{\dfrac{1}{2}, \; 128 \right\}$