Solve the equation: $\;$ $x^{2 \log x} = 10 x^2$
Given equation: $\;\;$ $x^{2 \log x} = 10 x^2$ $\;\;\; \cdots \; (1)$
Taking logarithims on both sides of equation $(1)$ gives
$\log \left(x^{2 \log x}\right) = \log \left(10 x^2\right)$
i.e. $\;$ $2 \log x \times \log x = \log 10 + \log x^2$
i.e. $\;$ 2 $\left(\log x\right)^2 = 1 + 2 \log x$
i.e. $\;$ $2 \left(\log x\right)^2 - 2 \log x - 1 = 0$
i.e. $\;$ $\log x = \dfrac{2 \pm \sqrt{4 + 8}}{4} = \dfrac{2 \pm \sqrt{12}}{4}$
i.e. $\;$ $\log x = \dfrac{2 \pm 2 \sqrt{3}}{4}$
i.e. $\;$ $\log x = \dfrac{1 \pm \sqrt{3}}{2}$
$\implies$ $x = 10^{\frac{1 \pm \sqrt{3}}{2}}$
i.e. $\;$ $x = \left(10^{1 \pm \sqrt{3}}\right)^{\frac{1}{2}}$
i.e. $\;$ $x = \sqrt{10^{1 \pm \sqrt{3}}}$
$\therefore \;$ The solution to the given equation is $\;\;$ $x = \left\{\sqrt{10^{1 + \sqrt{3}}}, \; \sqrt{10^{1 - \sqrt{3}}} \right\}$