Solve the equation: $\;$ $\log_2 \left(2x^2\right) \times \log_2 \left(16 x\right) = \dfrac{9}{2} \log^2_2 x$
Given equation: $\;\;$ $\log_2 \left(2x^2\right) \times \log_2 \left(16 x\right) = \dfrac{9}{2} \log^2_2 x$
i.e. $\;$ $\left(\log_2 2 + \log_2 x^2\right) \times \left(\log_2 16 + \log_2 x\right) = \dfrac{9}{2} \left(\log_2 x\right)^2$
i.e. $\;$ $\left(1 + 2 \log_2 x\right) \times \left(\log_2 2^4 + \log_2 x\right) = \dfrac{9}{2} \left(\log_2 x\right)^2$
i.e. $\;$ $\left(1 + 2 \log_2 x\right) \times \left(4 \log_2 2 + \log_2 x\right) = \dfrac{9}{2} \left(\log_2 x\right)^2$
i.e. $\;$ $\left(1 + 2 \log_2 x\right) \times \left(4 + \log_2 x\right) = \dfrac{9}{2} \left(\log_2 x\right)^2$ $\;\;\; \cdots \; (1)$
Let $\;$ $\log_2 x = p$ $\;\;\; \cdots \; (2)$
Then equation $(1)$ becomes
$\left(1 + 2p\right) \left(4 + p\right) = \dfrac{9}{2} p^2$
i.e. $\;$ $4 + 9p + 2p^2 = \dfrac{9}{2} p^2$
i.e. $\;$ $8 + 18 p + 4p^2 = 9 p^2$
i.e. $\;$ $5p^2 - 18p - 8 = 0$
i.e. $\;$ $\left(5p + 2\right) \left(p - 4\right) = 0$
i.e. $\;$ $5p + 2 = 0$ $\;$ or $\;$ $p - 4 = 0$
i.e. $\;$ $p = \dfrac{-2}{5}$, $\;$ or $\;$ $p = 4$
Substituting the value of $p$ in equation $(2)$ gives
when $\;$ $p = \dfrac{-2}{5}$, $\;$ we have from equation $(2)$
$\log_2 x = \dfrac{-2}{5}$ $\implies$ $x = 2^{\frac{-2}{5}}$
when $\;$ $p = 4$, $\;$ we have from equation $(2)$
$\log_2 x = 4$ $\implies$ $x = 2^4 = 16$
$\therefore \;$ The solution to the given equation is $\;\;$ $x = \left\{2^{\frac{-2}{5}}, \; 16 \right\}$