If the third term in the binomial expansion of 1+xlog2⁡x5equals 2560, then a possible value of x is

The (r + 1)th term in the expansion of (a + x)n is Given by  Tr+1=nCran−rxr.'. 3rd term in the expansion of 1+xlog2⁡x5 is  5C2(1)5−2xlog2⁡x2⇒ 5C2(1)5−2xlog2⁡x2⇒ 2560⇒ 10xlog2⁡x2=2560⇒  x2log2⁡x=256⇒ log2⁡x2log2⁡x=log2⁡256       (given) (taking Log2 on both sides) ⇒2log2⁡xlog2⁡x=8 ∵log2⁡256=log2⁡28=8log2⁡x2=4⇒log2⁡x=±2⇒log2⁡x=2 or log2⁡x=−2⇒x=4 or x=2−2=14