If the 7^{th} term in the binomial expansion | vedclass

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(B) The general term $T_{r+1}$ in the expansion of $(a+b)^n$ is given by $T_{r+1} = ^nC_r a^{n-r} b^r$.For the $7^{th}$ term,$r+1 = 7 \Rightarrow r =

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$e$

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(B) The general term $T_{r+1}$ in the expansion of $(a+b)^n$ is given by $T_{r+1} = ^nC_r a^{n-r} b^r$.For the $7^{th}$ term,$r+1 = 7 \Rightarrow r = 6$.Here,$n = 9$,$a = \frac{3}{\sqrt[3]{84}}$,and $b = \sqrt{3} \ln x$.$T_7 = ^9C_6 \left( \frac{3}{\sqrt[3]{84}} ight)^{9-6} (\sqrt{3} \ln x)^6 = 729$.$^9C_6 = ^9C_3 = \frac{9 	imes 8 	imes 7}{3 	imes 2 	imes 1} = 84$.$T_7 = 84 	imes \left( \frac{3}{\sqrt[3]{84}} ight)^3 	imes (\sqrt{3})^6 	imes (\ln x)^6 = 729$.$T_7 = 84 	imes \frac{27}{84} 	imes 27 	imes (\ln x)^6 = 729$.$27 	imes 27 	imes (\ln x)^6 = 729$.$729 	imes (\ln x)^6 = 729$.$(\ln x)^6 = 1$.Since $x > 0$,$\ln x = 1$ or $\ln x = -1$.If $\ln x = 1$,then $x = e$.If $\ln x = -1$,then $x = e^{-1} = \frac{1}{e}$.

If the $7^{th}$ 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

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