If the 7th term in the binomial expansion of $\left(\frac{3}{\sqrt[3]{84}}+\sqrt{3} \ln x\right)^9, x0$, is equal to 729 , then $x$ can be:

Question

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

Aicharya · Accepted Answer

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

If the 7th term in the binomial expansion of $\left(\frac{3}{\sqrt[3]{84}}+\sqrt{3} \ln x\right)^9, x>0$, is…

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