Solve the equation: $\;$ $\left[\sqrt{x}\right]^{\left(\log_5 x\right) - 1} = 5$
Given equation: $\;\;$ $\left[\sqrt{x}\right]^{\left(\log_5 x\right) - 1} = 5$
i.e. $\;$ $\left(\sqrt{x}\right)^{\log_5 x} \times \left(\sqrt{x}\right)^{-1} = 5$
i.e. $\;$ $\dfrac{\left(\sqrt{x}\right)^{\log_5 x}}{\sqrt{x}} = 5$ $\;\;\; \cdots \; (1)$
Taking logarithims to base $5$ on both sides of equation $(1)$ gives
$\log_5 \left[\dfrac{\left(\sqrt{x}\right)^{\log_5 x}}{\sqrt{x}}\right] = \log_5 5$
i.e. $\;$ $\log_5 \left(\sqrt{x}\right)^{\log_5 x} - \log_5 \sqrt{x} = 1$
i.e. $\;$ $\log_5 x \times \log_5 \sqrt{x} - \log_5 \sqrt{x} = 1$
i.e. $\;$ $\log_5 x \times \log_5 x^{\frac{1}{2}} - \log_5 x^{\frac{1}{2}} = 1$
i.e. $\;$ $\dfrac{1}{2} \log_5 x \times \log_5 x - \dfrac{1}{2} \log_5 x = 1$
i.e. $\;$ $\left(\log_5 x\right)^2 - \log_5 x - 2 = 0$
i.e. $\;$ $\left(\log_5 x\right)^2 - 2 \log_5 x + \log_5 x - 2 = 0$
i.e. $\;$ $\log_5 x \left(\log_5 x - 2\right) + 1 \left(\log_5 x - 2\right) = 0$
i.e. $\;$ $\left(\log_5 x + 1\right) \left(\log_5 x - 2\right) = 0$
i.e. $\;$ $\log_5 x = -1$ $\;$ or $\;$ $\log_5 x = 2$
$\implies$ $x = 5^{-1} = \dfrac{1}{5}$, $\;$ or $\;$ $x = 5^2 = 25$
$\therefore \;$ The solution to the given equation is $\;\;$ $x = \left\{\dfrac{1}{5}, \; 25 \right\}$