If the third term in the binomial expansion of 1 + x log 2 x 5 equals 2560 , then a possible value of x is

The general term in the expansion of ${\left(a+b\right)}^n$ is $T_{r+1}={}^{n}C_r{a}^{n-r}{b}^r.$

Given, in the expansion of ${\left(1+x^{{\log }_{2}x}\right)}^5,$ third term is $2560$

$\Rightarrow T_3={ }^5{C}_2{\left(x^{{\log }_{2}x}\right)}^2=2560$

Using, ${}^{n}C_r=\frac{n!}{r!·\left(n-r\right)!},$ we get

$\Rightarrow \frac{5!}{2!·3!}{\left(x^{{\log }_{2}x}\right)}^2=2560$

$\Rightarrow \frac{5·4·3!}{2\times 1·3!}\left(x^{2{\log }_{2}x}\right)=2560$

$\Rightarrow \left(x^{2{\log }_{2}x}\right)=256$

Taking logarithm to the base $2$ on both sides

$\Rightarrow {\log }_2\left(x^{2{\log }_{2}x}\right)={\log }_{2}256$

Now, using ${\log }_a{m}^n=n{\log }_{a}m \& {\log }_{a}a=1,$

$\Rightarrow 2lo{g}_{2}x\left({\log }_{2}x\right)={\log }_2{2}^8$

$\Rightarrow 2{\left({\log }_{2}x\right)}^2=8$

$\Rightarrow \left({\log }_{2}x\right)=\pm 2$

Using, ${\log }_{a}x=b, \Rightarrow x=a^b$

$x=2^2, 2^{-2}$

$\Rightarrow x=4, \frac{1}{4}$

Here, only $x=\frac{1}{4}$ is given in the options.