If the third term in the binomial expansion of  

${\left(1+2^{{\log }_{2}x}\right)}^5$ is equal to 2560, then a possible value of  $x$ is 

The third  term  $T_3$ in the binomial expansion of 

 ${\left(1+x^{{\log }_{2}x}\right)}^5$  is given by 

      $T_3{=}^5{C}_2{\left(x^{{\log }_{2}x}\right)}^2=10{\left({\log }_{2}x\right)}^2$

It is given that  $T_3=2560$ 

  $\begin{array}{l}\therefore 10{\left(x^{{\log }_{2}x}\right)}^2=2560\Rightarrow {\left(x^{{\log }_{2}x}\right)}^2=256\Rightarrow x^{{\log }_{2}x}=16\\ \Rightarrow {\log }_{2}x={\log }_{x}16\Rightarrow {\log }_{2}x=\frac{4}{{\log }_{2}x}\\ \Rightarrow {\left({\log }_{2}x\right)}^2=4\Rightarrow {\log }_{2}x=\pm 2\Rightarrow x=4,\frac{1}{4}\end{array}$