Solve the equation: $\;$ $\log^2 \left(100x\right) - \log^2 \left(10 x\right) + \log^2 x = 6$
Given equation: $\;\;$ $\log^2 \left(100x\right) - \log^2 \left(10 x\right) + \log^2 x = 6$
i.e. $\;$ $\left(\log 100 + \log x\right)^2 - \left(\log 10 + \log x\right)^2 + \log^2 x = 6$
i.e. $\;$ $\left(2 + \log x\right)^2 - \left(1 + \log x\right)^2 + \log^2 x = 6$
i.e. $\;$ $\log^2 x + 4 \log x + 4 - \left(\log^2 x + 2 \log x + 1\right) + \log^2 x = 6$
i.e. $\;$ $\log^2 x + 2 \log x - 3 = 0$
i.e. $\;$ $\log^2 x + 3 \log x - \log x - 3 = 0$
i.e. $\;$ $\log x \left(\log x + 3\right) - 1 \left(\log x + 3\right) = 0$
i.e. $\;$ $\left(\log x + 3\right) \left(\log x - 1\right) = 0$
i.e. $\;$ $\log x + 3 = 0$ $\;$ or $\;$ $\log x - 1 = 0$
i.e. $\;$ $\log x = -3$ $\;$ or $\;$ $\log x = 1$
i.e. $\;$ $x = 10^{-3}$ $\;$ or $\;$ $x = 10^1 = 10$
$\therefore \;$ The solution to the given equation is $\;\;$ $x = \left\{10^{-3}, \; 10 \right\}$