Question

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

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

Answer 1

The correct option is A $8^2$

$(\frac{2}{x}+x^{{\log }_{8}x}{)}^6$ will have 7 terms.
$\therefore 4^{th}term={}^6{C}_3\cdot \left(\frac{2}{x}{)}^3\right(x^{{\log }_{8}x}{)}^3\Rightarrow 20\times 8^7=\frac{20\times 8}{x^3}\cdot x^{3{\log }_{8}x}\Rightarrow 8^6=\frac{x^{{\log }_{2}x}}{x^3}=x^{{\log }_{2}x-3}\Rightarrow 2^{18}=x^{{\log }_{2}x-3}$
Taking log on both sides, we get
$18=({\log }_{2}x-3){\log }_{2}x$
Let ${\log }_{2}x=t$
$\Rightarrow t^2-3t-18=0\Rightarrow t=6,t=-3If{\log }_{2}x=6\Rightarrow x=2^6=8^{2}andif{\log }_{2}x=-3\Rightarrow x=2^{-3}=8^{-1}$