Solve the equation: $\;$ $x^{\left(\log x\right) + 1} = 10^6$
Given equation: $\;\;$ $x^{\left(\log x\right) + 1} = 10^6$
i.e. $\;$ $x^{\log x} \times x^1 = 10^6$ $\;\;\; \cdots \; (1)$
Taking logarithims on both sides of equation $(1)$ gives
$\log \left(x^{\log x} \times x\right) = \log {10}^6$
i.e. $\;$ $\log \left(x^{\log x}\right) + \log x = 6 \log 10$
i.e. $\;$ $\log x \times \log x + \log x = 6$
i.e. $\;$ $\left(\log x\right)^2 + \log x - 6 = 0$
i.e. $\;$ $\left(\log x\right)^2 + 3 \log x - 2 \log x - 6 = 0$
i.e. $\;$ $\log x \left(\log x + 3\right) - 2 \left(\log x + 3\right) = 0$
i.e. $\;$ $\left(\log x + 3\right) \left(\log x - 2\right) = 0$
i.e. $\;$ $\log x = -3$ $\;$ or $\;$ $\log x = 2$
$\implies$ $x = 10^{-3}$ $\;$ or $x = 10^2$
$\therefore \;$ The solution to the given equation is $\;\;$ $x = \left\{10^{-3}, 10^2 \right\}$