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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:
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Verified Answer
The correct answer is:
$e$
$e$
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
$$
\begin{aligned}
& 729={ }^9 \mathrm{C}_6\left(\frac{3}{\sqrt[3]{84}}\right)^3 \cdot(\sqrt{3} \ln x)^6 \\
& \Rightarrow 3^6=84 \times \frac{3^3}{84} \times 3^3 \times(6 \ln x) \\
& \Rightarrow(\ln x)^6=1 \Rightarrow(\ln x)^6=(\ln e)^6 \\
& \Rightarrow x=\mathrm{e}
\end{aligned}
$$
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
$$
\begin{aligned}
& 729={ }^9 \mathrm{C}_6\left(\frac{3}{\sqrt[3]{84}}\right)^3 \cdot(\sqrt{3} \ln x)^6 \\
& \Rightarrow 3^6=84 \times \frac{3^3}{84} \times 3^3 \times(6 \ln x) \\
& \Rightarrow(\ln x)^6=1 \Rightarrow(\ln x)^6=(\ln e)^6 \\
& \Rightarrow x=\mathrm{e}
\end{aligned}
$$
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