If the 7th term in the binomial expansion of 3841/3+3ln⁡x9, (x>0) is equal to 729, then x is equal to

A
$e, 1 / e$
B
$2 e, 1 / 2 e$
C
$e^{2}, e^{- 2}$
D
$1 / e$

Solution:

$T_{7} =^{9} C_{6} {(\frac{3}{84^{1 / 3}})}^{9 - 6} (\sqrt{3} \ln x)^{6}$
$\begin{array}{ll} \Rightarrow \frac{9 \times 8 \times 7}{3 \times 2} \cdot \frac{3^{3}}{84} 3^{3} (\ln x)^{6} = 729 \\ \Rightarrow (\ln x)^{6} = 1 \\ \Rightarrow \ln x = \pm 1 \Rightarrow x = e, 1 / e \end{array}$

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