Solve the equation: $\;$ $x^{\frac{\log x + 5}{3}} = 10^{5 + \log x}$
Given equation: $\;\;$ $x^{\frac{\log x + 5}{3}} = 10^{5 + \log x}$ $\;\;\; \cdots \; (1)$
Taking logarithims on both sides of equation $(1)$ gives
$\log \left(x^{\frac{\log x + 5}{3}}\right) = \log \left(10^{5 + \log x}\right)$
i.e. $\;$ $\left(\dfrac{\log x + 5}{3}\right) \log x = \left(5 + \log x\right) \log 10$
i.e. $\;$ $\dfrac{1}{3} \left(\log x\right)^2 + \dfrac{5}{3} \log x = 5 + \log x$
i.e. $\;$ $\dfrac{1}{3} \left(\log x\right)^2 + \dfrac{2}{3} \log x - 5 = 0$
i.e. $\;$ $\left(\log x\right)^2 + 2 \log x - 15 = 0$
i.e. $\;$ $\left(\log x\right)^2 + 5 \log x - 3 \log x - 15 = 0$
i.e. $\;$ $\log x \left(\log x + 5\right) - 3 \left(\log x + 5\right) = 0$
i.e. $\;$ $\left(\log x - 3\right) \left(\log x + 5\right) = 0$
i.e. $\;$ $\log x = 3$ $\;\;$ or $\;\;$ $\log x = -5$
$\implies$ $x = 10^3$ $\;\;$ or $\;\;$ $x = 10^{-5}$
$\therefore \;$ The solution to the given equation is $\;\;$ $x = \left\{10^{-5}, \; 10^3 \right\}$