If $ n=(1999) $ ! then $ \sum_\limits {x=1}^{1999} \log _{n} x $ is equal to

JEE MATHEMATICS MCQ: SEQUENCE OF SERIES

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Application +4 | -1

n=(1999)

! then

\sum\limits_{{x=1}}^{1999} \log _{n} x

is equal to

View answer and explanation

Correct option: A

Option A (1) is correct because the sum of logarithms with the same base is the logarithm of the product, which simplifies to 1.

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A-(B \cap C)

(B \cup C)-A

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{x^2},{b^2},{y^2}

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\frac{3}{4}