Solve the equation: $\;$ $x^{\frac{\left(\log x\right) + 7}{4}} = 10^{\left(\log x\right) + 1}$
Given equation: $\;\;$ $x^{\frac{\left(\log x\right) + 7}{4}} = 10^{\left(\log x\right) + 1}$ $\;\;\; \cdots \; (1)$
Taking logarithims on both sides of equation $(1)$ gives
$\log \left(x^{\frac{\log x + 7}{4}}\right) = \log \left(10^{\log x + 1}\right)$
i.e. $\;$ $\left(\dfrac{\log x + 7}{4}\right) \times \log x = \left(\log x + 1\right) \times \log 10$
i.e. $\;$ $\dfrac{\left(\log x\right)^2 + 7 \log x}{4} = \left(\log x + 1\right) \times 1$
i.e. $\;$ $\left(\log x\right)^2 + 7 \log x = 4 \log x + 4$
i.e. $\;$ $\left(\log x\right)^2 + 3 \log x - 4 = 0$
i.e. $\;$ $\left(\log x\right)^2 + 4 \log x - \log x - 4 = 0$
i.e. $\;$ $\log x \left(\log x + 4\right) - 1 \left(\log x + 4\right) = 0$
i.e. $\;$ $\left(\log x - 1\right) \left(\log x + 4\right) = 0$
i.e. $\;$ $\log x = 1$ $\;$ or $\;$ $\log x = -4$
$\implies$ $x = 10^1 = 10$ $\;$ or $x = 10^{-4}$
$\therefore \;$ The solution to the given equation is $\;\;$ $x = \left\{10^{-4}, \; 10 \right\}$