If n=(1999) ! , then ∑_x=1^1999 log _n x is equal to

Correct answer: (A) 1. ∑_x=1^1999 log _n x =log _(1999) ! 1+log _(1999) ! 2+…+log _(1999) ! 1999 =log _(1999) !(1.2 .3 … 1999) =log _(1999) !(1999) !=1

Q73 / 120 · VITEEE 2013 · Mathematics

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

Correct answer
A
1
B
0
C
$\sqrt[1999]{1999}$
D
$-1$

Solution

$\sum_{x=1}^{1999} \log _{n} x$ $=\log _{(1999) !} 1+\log _{(1999) !} 2+\ldots+\log _{(1999) !} 1999$ $=\log _{(1999) !}(1.2 .3 \ldots 1999)$ $=\log _{(1999) !}(1999) !=1$

Hence, the correct answer is option: 'A'.

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

Solution

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