Solve the equation: $\;$ $\log^2 \left(100 x\right) + \log^2 \left(10 x\right) = 14 + \log \left(\dfrac{1}{x}\right)$
Given equation: $\;\;$ $\log^2 \left(100 x\right) + \log^2 \left(10 x\right) = 14 + \log \left(\dfrac{1}{x}\right)$
i.e. $\;$ $\left(\log 100 + \log x\right)^2 + \left(\log 10 + \log x\right)^2 = 14 + \log 1 - \log x$
i.e. $\;$ $\left(\log 10^2 + \log x\right)^2 + \left(1 + \log x\right)^2 = 14 + 0 - \log x$
i.e. $\;$ $\left(2 \log 10 + \log x\right)^2 + \left(1 + \log x\right)^2 = 14 - \log x$
i.e. $\;$ $\left(2 + \log x\right)^2 + \left(1 + \log x\right)^2 = 14 - \log x$
i.e. $\;$ $4 + \log^2 x + 4 \log x + 1 + \log^2 x + 2 \log x = 14 - \log x$
i.e. $\;$ $2 \log^2 x + 7 \log x - 9 = 0$
i.e. $\;$ $2 \log^2 x + 9 \log x - 2 \log x - 9 = 0$
i.e. $\;$ $2 \log x \left(\log x - 1\right) + 9 \left(\log x - 1\right) = 0$
i.e. $\;$ $\left(2 \log x + 9\right) \left(\log x - 1\right) = 0$
i.e. $\;$ $\log x = \dfrac{-9}{2}$ $\;$ or $\;$ $\log x = 1$
$\implies$ $x = 10^{\frac{-9}{2}} = \sqrt{10^{-9}}$ $\;$ or $\;$ $x = 10^1 = 10$
$\therefore \;$ The solution to the given equation is $\;\;$ $x = \left\{\sqrt{10^{-9}}, \; 10 \right\}$