If the fourth term in the binomial expansion of ((2/x) + xlog8 x)6 (x 0) is 20 × 87, then a value of x is :

Question

If the fourth term in the binomial expansion of ((2/x) + xlog8 x)6 (x > 0) is 20 × 87, then a value of x is : (A) 8 (B) 82 (C) 8-2 (D) 83

Tardigrade Admin · Accepted Answer

T4 = T3+1 = (63) ((2/x))3 . (xlogg x)3 20 × 87 = (160/x3). x3 logg x 86 = xlog2 x -3 218 = xlog2 x-3 ⇒ 18 = (log2 x- 3) (log2 x) Let log2 x = t ⇒ t2 - 3t - 18 = 0 ⇒ (t - 6)(t + 3) = 0 ⇒ t = 6, - 3 log2 x = 6 ⇒ x = 26 = 82 log2 x = -3 ⇒ x = 2-3 = 8-1

Q. If the fourth term in the binomial expansion of $(\frac{2}{x} + x^{l o g_{8} x})^{6} (x > 0) i s 20 \times 8^{7}$ , then a value of $x$ is :

$8$

$8^{2}$

$8^{- 2}$

$8^{3}$

Q. If the fourth term in the binomial expansion of (x2​+xlog8​x)6(x>0)is20×87, then a value of x is :

8

82

8−2

83

T4​=T3+1​=(36​)(x2​)3.(xlogg​x)3 20×87=x3160​.x3logg​x 86=xlog2​x−3 218=xlog2​x−3 ⇒18=(log2​x−3)(log2​x) Let log2​ x = t ⇒t2−3t−18=0 ⇒ (t - 6)(t + 3) = 0 ⇒ t = 6, - 3 log2​x=6⇒x=26=82 log2​x=−3⇒x=2−3=8−1

Q. If the fourth term in the binomial expansion of $(\frac{2}{x} + x^{l o g_{8} x})^{6} (x > 0) i s 20 \times 8^{7}$ , then a value of $x$ is :

$8$

$8^{2}$

$8^{- 2}$

$8^{3}$