If the third term in the binomial expansion of (1+x^ log_2x)^5 equals 2560, then a possible value of x is

The general term in the expansion of ${(a+b)}^{n}$(a+b)n is ${T}_{r+1}=Crn{a}^{n-r}{b}^{r}.$Tr+1​=Crnan−rbr.

Given, in the expansion of ${(1+{x}^{{\mathrm{log}}_{2}x})}^{5},$(1+xlog2​x)5, third term is $2560$2560

$$\Rightarrow {T}_{3}={}^{5}{C}_{2}{({x}^{{\mathrm{log}}_{2}x})}^{2}=2560$$⇒T3​=5C2​(xlog2​x)2=2560

Using, $Crn=\frac{n!}{r!\cdot (n-r)!},$Crn=r!⋅(n−r)!n!​, we get

$$\Rightarrow \frac{5!}{2!\cdot 3!}{({x}^{{\mathrm{log}}_{2}x})}^{2}=2560$$⇒2!⋅3!5!​(xlog2​x)2=2560

$$\Rightarrow \frac{5\cdot 4\cdot 3!}{2\times 1\cdot 3!}({x}^{2{\mathrm{log}}_{2}x})=2560$$⇒2×1⋅3!5⋅4⋅3!​(x2log2​x)=2560

$$\Rightarrow ({x}^{2{\mathrm{log}}_{2}x})=256$$⇒(x2log2​x)=256

Taking logarithm to the base $2$2 on both sides

$$\Rightarrow {\mathrm{log}}_{2}({x}^{2{\mathrm{log}}_{2}x})={\mathrm{log}}_{2}256$$⇒log2​(x2log2​x)=log2​256

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

$$\Rightarrow 2lo{g}_{2}x({\mathrm{log}}_{2}x)={\mathrm{log}}_{2}{2}^{8}$$⇒2log2​x(log2​x)=log2​28

$$\Rightarrow 2{({\mathrm{log}}_{2}x)}^{2}=8$$⇒2(log2​x)2=8

$$\Rightarrow ({\mathrm{log}}_{2}x)=\pm 2$$⇒(log2​x)=±2

Using, ${\mathrm{log}}_{a}x=b,\Rightarrow x={a}^{b}$loga​x=b,⇒x=ab

$$x={2}^{2},{2}^{-2}$$x=22,2−2

$$\Rightarrow x=4,\frac{1}{4}$$⇒x=4,41​

Here, only $x=\frac{1}{4}$x=41​ is given in the options.