If n= 1999 ! then \large \sf\sum\limits_(x = 1)^(1999) log_(n)(x) is equal to a) -1 b) 0 c) 1 d) \sf \sqrt[1999]{1999}

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If n= 1999 ! then
$\large \sf\sum\limits_(x = 1)^(1999) log_(n)(x)$ is equal to

a) -1
b) 0
c) 1
d)
$\sf \sqrt[1999]{1999}$

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20 votes

Answer:

a) 1

Explanation:

added in the picture

$If n= 1999 ! then \large \sf\sum\limits_(x = 1)^(1999) log_(n)(x) is equal to a) -1 b-example-1$

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