Question

If the fourth term in the binomial expansion of ${(\frac{2}{x}+x^{lo{g}_{5}x})}^6(x>0)$ is $20\times 8^7$, then a value of x is?

• A. $8$
• B. $8^{-1}$
• C. $8^{-2}$
• D. $8^3$

Answer 1

The correct option is B $8^{-1}$

$T_4=T_{3+1}=\left(\begin{array}{c}6\\ 3\end{array}\right){\left(\frac{2}{x}\right)}^3\cdot (x^{lo{g}_{5}x}{)}^3$

$20\times 8^7=\frac{160}{x^3}\cdot x^{3lo{g}_{5}x}$

$8^6=x^{lo{g}_{2}x-3}$

$2^{18}=x^{lo{g}_{2}x-3}$

$\Rightarrow 18=(lo{g}_{2}x-3)\left(lo{g}_{2}x\right)$

Let $lo{g}_{2}x=t$

$\Rightarrow t^2-3t-18=0$

$\Rightarrow (t-6)(t+3)=0$

$\Rightarrow t=6,-3$

$lo{g}_{2}x=6$

$\Rightarrow x=2^6=8^2$

${\log }_{2}x=-3$

$\Rightarrow x=2^{-3}=8^{-1}$.