If the middle term in the expansion of ${\left(\frac{1}{x}+x^{{\log }_{2}x}\right)}^5$ is 40 then $x$equal 

The middle term of ${\left(\frac{1}{x}+x^{{\log }_{2}x}\right)}^5$ is

$\begin{array}{l}{ }^5{C}_2{\left(\frac{1}{x}\right)}^3{\left(x^{{\log }_{2}x}\right)}^2=10x^{2{\log }_{2}x-3}=40\\ \therefore x^{2{\log }_{2}x-3}=4=2^2\end{array}$

Taking logarithm to the base 2 of both the sides, we get

$\begin{array}{r}(2t-3)t=2\text{ where }t={\log }_{2}x\\ \Rightarrow 2t^2-3t-2=0\text{ or }(2t+1)(t-2)=0\\ \Rightarrow t=-1/2\text{ or }2\Rightarrow x=1/\sqrt{2},4\end{array}$