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

Step-by-step solution

The general term in the expansion of a + b n is T r + 1 = C r n a n - r b r . Given, in the expansion of 1 + x log 2 x 5 , third term is 2560 ⇒ T 3 =   5 C 2 x log 2 x 2 = 2560 Using, C r n = n ! r ! · n - r ! , we get ⇒ 5 ! 2 ! · 3 ! x log 2 x 2 = 2560 ⇒ 5 · 4 · 3 ! 2 × 1 · 3 ! x 2 log 2 x = 2560 ⇒ x 2 log 2 x = 256 Taking logarithm to the base 2 on both sides ⇒ log 2 x 2 log 2 x = log 2 256 Now, using log a m n = n log a m   &   log a