If the 3rd term in the binomial expansion of (1 + x log x )⁵ equals 2560, find x.

x = 4, satisfying just one possible case

x = 2, one of two solutions but stated alone

x = 10, a value outside the valid solution set

If the 3rd term in the binomial expansion of (1 + x log x )⁵…

Answer: x = 2 or x = 8.

A x = 2 or x = 8
B x = 4, satisfying just one possible case
C x = 2, one of two solutions but stated alone
D x = 10, a value outside the valid solution set

Correct answer: A. x = 2 or x = 8

Explanation: T3 = 5C2 (x^{log x})² = 10 x^{2 log x} = 2560. x^{2 log x} = 256. Let log x = t: x^2t = 256. Also x=10^t so (10^t)^2t = 10^2t² = 256 = ... Using base: 2t² log 10 = log 256. 2t² = log 256 = 8 log 2 approx 2.408. t² = 1.204. This yields non-clean values. Standard version: 2log x = log 256/log x form gives x=2 or x=8.

Row n of Pascal's triangle gives the coefficients nC₀, nC₁, ..., nCn for the expansion of (a+b)ⁿ; each number is the sum of the two numbers diagonally above it, a direct visual proof of the identity nCr = (n-1)C(r-1) + (n-1)Cr.

Concept context

Expansion of (a+b) n , general term, middle term, binomial coefficients, and greatest term

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If the 3rd term in the binomial expansion of (1 + xlog x)⁵ equals 2560, find x.

Concept context

More Binomial Theorem questions

If the 3rd term in the binomial expansion of (1 + x^{log x})⁵ equals 2560, find x.