Solve the equation: $\;$ $x^{3 \log x - \frac{1}{\log x}} = \sqrt[3]{10}$
Given equation: $\;\;$ $x^{3 \log x - \frac{1}{\log x}} = \sqrt[3]{10}$
i.e. $\;$ $\left(x^{3 \log x - \frac{1}{\log x}}\right)^3 = \left(10^{\frac{1}{3}}\right)^3$
i.e. $\;$ $x^{9 \log x - \frac{3}{\log x}} = 10$ $\;\;\; \cdots \; (1)$
Taking logarithim to base $10$ on both sides of equation $(1)$ gives
i.e. $\;$ $\log \left(x^{9 \log x - \frac{3}{\log x}}\right) = \log 10$
i.e. $\;$ $\left(9 \log x - \frac{3}{\log x}\right) \log x = 1$
i.e. $\;$ $9 \left(\log x\right)^2 - 3 = 1$
i.e. $\;$ $9 \left(\log x\right)^2 = 4$
i.e. $\;$ $3 \log x = \pm 2$
i.e. $\;$ $\log x = \dfrac{2}{3}$, $\;$ or $\;$ $\log x = \dfrac{-2}{3}$
$\implies$ $x = 10^{\frac{2}{3}}$ $\;$ or $\;$ $x = 10^{\frac{-2}{3}}$
$\therefore \;$ The solution to the given equation is $\;\;$ $x = \left\{10^{\frac{-2}{3}}, \; 10^{\frac{2}{3}} \right\}$