Question

If the fourth term in the Binomial expansion of $\left(\frac{2}{x}+x^{\log_{8}x}\right)^{6}$, $x>0$ is $20\times 8^{7}$, then a value of $x$ is

• A. $8^{3}$
• B. 8
• C. $8^{-2}$
• D. $8^{2}$

Answer 1

Answer: (D)

$8^2$

Answer 2

The $(r+1)$-th term in the binomial expansion of $(a+b)^n$ is $T_{r+1} = \binom{n}{r} a^{n-r} b^r$. For the expansion $\left(\frac{2}{x}+x^{\log_{8}x}\right)^{6}$, $n=6$, $a=\frac{2}{x}$, $b=x^{\log_{8}x}$. The fourth term ($T_4$) corresponds to $r=3$. $T_4 = \binom{6}{3} \left(\frac{2}{x}\right)^{6-3} \left(x^{\log_{8}x}\right)^{3} = 20 \left(\frac{2}{x}\right)^{3} x^{3\log_{8}x} = 20 \frac{8}{x^3} x^{3\log_{8}x} = 160 x^{-3+3\log_{8}x}$. Given $T_4 = 20 \times 8^7$. $160 x^{-3+3\log_{8}x} = 20 \times 8^7$ $8 x^{-3+3\log_{8}x} = 8^7$ $x^{-3+3\log_{8}x} = 8^6$. Let $y = \log_{8}x$, so $x = 8^y$. $(8^y)^{-3+3y} = 8^6 \implies 8^{y(-3+3y)} = 8^6$. Equating exponents: $-3y+3y^2 = 6 \implies 3y^2-3y-6=0 \implies y^2-y-2=0$. Factoring gives $(y-2)(y+1)=0$, so $y=2$ or $y=-1$. If $y=2$, $\log_{8}x = 2 \implies x=8^2$. If $y=-1$, $\log_{8}x = -1 \implies x=8^{-1}$. Since $8^2$ is an option, it is a valid solution.