Solve the equation: $\;$ $x^{1 + \log x} = 10x$
Given equation: $\;\;$ $x^{1 + \log x} = 10x$
i.e. $\;$ $1 + \log x = \log_x {10x}$
i.e. $\;$ $1 + \log x = \log_x {10} + \log_x x$ $\;\;\;$ $\left[\because \; \log_a \left(CD\right) = \log_a C + \log_a D\right]$
i.e. $\;$ $1 + \log x = \dfrac{\log_{10} 10}{\log_{10} x} + 1$ $\;\;\;$ $\left[\because \; \log_b a = \dfrac{\log_m a}{\log_m b}; \;\; \log_a a = 1\right]$
i.e. $\;$ $\log_{10} x = \dfrac{1}{\log_{10} x}$
i.e. $\;$ $\left(\log_{10} x\right)^2 = 1$
i.e. $\;$ $\log_{10} x = \pm 1$
Now, $\;$ $\log_{10} x = 1$ $\implies$ $x = 10^1 = 10$
and $\;$ $\log_{10} x = -1$ $\implies$ $x = 10^{-1} = \dfrac{1}{10} = 0.1$
$\therefore \;$ The solution to the given equation is $\;\;$ $x = \left\{0.1, \; 10 \right\}$