Solve the equation: $\;$ $\left(\dfrac{1}{5}\right)^{\log^2 x - \log x} = \dfrac{1}{125} \times 5^{\log x - 1}$
Given equation: $\;\;$ $\left(\dfrac{1}{5}\right)^{\log^2 x - \log x} = \dfrac{1}{125} \times 5^{\log x - 1}$
i.e. $\;$ $\left(\dfrac{1}{5}\right)^{\log^2 x - \log x} = \dfrac{1}{5^3} \times 5^{\log x - 1}$
i.e. $\;$ $\left[\left(5^{-1}\right)\right]^{\log^2 x - \log x} = 5^{\log x - 4}$
i.e. $\;$ $5^{\log x - \log^2 x} = 5^{\log x - 4}$
i.e. $\;$ $\log x - \log^2 x = \log x - 4$
i.e. $\;$ $\log^2 x = 4$
i.e. $\;$ $\log x = 2$ $\;$ or $\;$ $\log x = -2$
i.e. $\;$ $x = 10^2$ $\;$ or $x = 10^{-2}$
$\therefore \;$ The solution to the given equation is $\;\;$ $x = \left\{10^2, 10^{-2} \right\}$