If the fourth term in the Binomial expansion of (x2+xlog8x)6(x>0) is 20×87, then a value of x is:
Solutions · 1
Official solution
\[T_4=\binom63\left(\frac2x\right)^3\left(x^{\log_8x}\right)^3=20\times\frac{8}{x^3}\times x^{3\log_8x}=20\times8\times x^{3\log_8x-3}\]
Given =20×87: x3log8x−3=86=x18log8x-type relation. Taking log8 on both sides (following source): (log8x)2−log8x=2⇒log8x=−1 or log8x=2
\[x=8^{-1}\text{ or }x=8^2\]