Solve the equation: $\;$ $\log^2 x - 3 \log x = \log x^2 - 4$
Given equation: $\;\;$ $\log^2 x - 3 \log x = \log x^2 - 4$
i.e. $\;$ $\log^2 x - 3 \log x = 2 \log x - 4$
i.e. $\;$ $\left(\log x\right)^2 - 5 \log x + 4 = 0$
i.e. $\;$ $\left(\log x\right)^2 - 4 \log x - \log x + 4 = 0$
i.e. $\;$ $\log x \left(\log x - 4\right) - 1 \left(\log x - 4\right) = 0$
i.e. $\;$ $\left(\log x - 1\right) \left(\log x - 4\right) = 0$
i.e. $\;$ $\log x = 1$ $\;$ or $\;$ $\log x = 4$
$\implies$ $x = 10^1 = 10$ $\;$ or $\;$ $x = 10^4$
$\therefore \;$ The solution to the given equation is $\;\;$ $x = \left\{10, \; 10^4 \right\}$