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If $\log_{3} x + \log_{9} x + \log_{27} x + \log_{729} x = 4$ , then $x =$

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answer is B.

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Detailed Solution

$\log_{3} x + \log_{9} x + \log_{27} x + \log_{729} x = 4$

$\log_{3} x (1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{6}) = 4$

$\log_{3} x . \frac{12}{6} = 4 \Rightarrow x = 9$

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