Question

If the $7th$ term in the binomial expansion of ${\left(\frac{3}{\sqrt[3]{84}}+\sqrt{3}Inx\right)}^9,x>0$, is equal to $729$, then $x$ can be:

• A. $e^2$
• B. $e$
• C. $\frac{e}{2}$
• D. $2e$

Answer 1

Answer: (B)

$Letr+1=7\Rightarrow r=6$
Given expansion is
${\left(\frac{3}{\sqrt[3]{84}}+\sqrt{3}lnx\right)}^{{}^9},x>0$
We have
$T_{r+1}{=}^n{C}_r{\left(x\right)}^{n-r}{a}^{r}for{\left(x+a\right)}^n.$
$\therefore$ According to the question
$729{=}^9{C}_6{\left(\frac{3}{\sqrt[3]{84}}\right)}^{{}^3}.{\left(\sqrt{3}lnx\right)}^6$
$\Rightarrow 3^6=84\times \frac{3^3}{84}\times 3^3\times \left(6lnx\right)$
$\Rightarrow {\left(lnx\right)}^6=1\Rightarrow {\left(lnx\right)}^6={\left(lne\right)}^6$
$\Rightarrow x=e$