If the 7^{th} term from the beginning in the | 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,$r = 6$ and $n = 9$.$T_

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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,$r = 6$ and $n = 9$.$T_7 = {^9C_6} \left( \frac{3}{(84)^{1/3}} ight)^{9-6} (\sqrt{3} \ln x)^6 = 729$.$T_7 = {^9C_3} \left( \frac{3^3}{84} ight) (3^3) (\ln x)^6 = 729$.$T_7 = 84 	imes \frac{27}{84} 	imes 27 	imes (\ln x)^6 = 729$.$27 	imes 27 	imes (\ln x)^6 = 729$.$729 	imes (\ln x)^6 = 729$.$(\ln x)^6 = 1$.$\ln x = \pm 1$.If $\ln x = 1$,then $x = e$.If $\ln x = -1$,then $x = \frac{1}{e}$.Given the options,$x = e$ is the correct value.

If the $7^{th}$ term from the beginning in the binomial expansion of ${\left( {\frac{3}{{{{\left( {84} \right)}^{\frac{1}{3}}}}} + \sqrt 3 \ln x} \right)^9}$ for $x > 0$ is equal to $729$,then the possible value of $x$ is:

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