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

Question

Accepted Answer

Correct answer: A. (a) Third term of (1+x^ log_2x)^5=5_C_2 (x^ log_2x )^5-3 =5_C_2 (x^ log_2x )^2 Given, 5_C_2 (x^ log_2x )^2=2560 ⇒ (x^ log_2x )^2=256=( 16)^2 ⇒x^ log_2x=16 or x^ log_2x=-16 (rejected) ⇒x^ log_2x=16 ⇒ log_2x log_2x= log_216=4 ⇒ log_2x= 2 ⇒x=2^2 or 2^-2 ⇒x=4 or 14

$In the third term in binomial expansion of (1 + x^{log_{2} x})^{5} equals 2560, then a possible value of x is :$

Solution

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