Mathematics - Binomial Theorem Question with Solution | TestHub

Let $r+1=7\Rightarrow r=6$ Given expansion is $${\left(\frac{3}{\sqrt[3]{84}}+\sqrt{3}\ln x\right)}^9,x>0$$ We have $${\mathrm{T}}_{r+1}={}^n{\mathrm{C}}_r(x{)}^{n-r}{a}^r\text{ for }(x+a{)}^n\text{. }$$ $\therefore$ According to the question $$\begin{array}{rl}& 729={}^9{\mathrm{C}}_6{\left(\frac{3}{\sqrt[3]{84}}\right)}^3\cdot (\sqrt{3}\ln x{)}^6\\ & \Rightarrow 3^6=84\times \frac{3^3}{84}\times 3^3\times (6\ln x)\\ & \Rightarrow (\ln x{)}^6=1\Rightarrow (\ln x{)}^6=(\ln e{)}^6\\ & \Rightarrow x=\mathrm{e}\end{array}$$