jmc

Applying the logarithmic identities ${\log }_a{b}^c=c{\log }_{a}b$ and ${\log }_{a^c}b=(1/c){\log }_{a}b$, we find $$\begin{array}{rl}10& ={\log }_{4}x+{\log }_2{x}^2\\ & ={\log }_{4}x+2{\log }_{2}x\\ & ={\log }_{2^2}x+2{\log }_{2}x\\ & =\frac{1}{2}{\log }_{2}x+2{\log }_{2}x\\ & =\frac{5}{2}{\log }_{2}x.\end{array}$$Therefore, ${\log }_{2}x=4$, which implies $x=2^4=\boxed{16}$.