Question

If $ n = (1999)!, $ then $ \sum\limits_{x = 1}^{1999} {{{\log }_n}x} $ is equal to
A. $ 1 $
B. $ 0 $
C. $ \sqrt[9]{{1999}} $
D. $ - 1 $

Answer 1

Hint: Here we are going to use the laws of the logarithmic product rule and the base of the quotient rule. Follow the step by step approach and simplify the intermediate equations carefully.
$ 1:{\log _b}a = \dfrac{{\log a}}{{\log b}} $
$ 2:\log {a_1} + \log {a_2} + \log {a_3}......\log {a_n} = \log ({a_1}{a_2}{a_3}....{a_n}) $

Complete step-by-step answer:
Here, we are going to use the properties of the logarithm
Property $ 1:{\log _b}a = \dfrac{{\log a}}{{\log b}} $
Property $ 2:\log {a_1} + \log {a_2} + \log {a_3}......\log {a_n} = \log ({a_1}{a_2}{a_3}....{a_n}) $
Now, take the given relation – $ \sum\limits_{x = 1}^{1999} {{{\log }_n}x} $
By using the property $ 1 $ ,
 $ \sum\limits_{x = 1}^{1999} {{{\log }_n}x} = \sum\limits_{x = 1}^{1999} {\dfrac{{\log x}}{{\log n}}} $
Since, the value of “n” is fixed and the constant number. Take common from the above equation.
 $ \sum\limits_{x = 1}^{1999} {{{\log }_n}x} = \dfrac{1}{{\log n}}\sum\limits_{x = 1}^{1999} {\log x} $
Place the value of “n” and “x”
 $ \sum\limits_{x = 1}^{1999} {{{\log }_n}x} = \dfrac{1}{{\log (1999!)}}[\log 1 + \log 2 + ....\log 1999] $
Now, using the property of the logarithm, the sum of the logs can be written as the product of the log.
 $ \sum\limits_{x = 1}^{1999} {{{\log }_n}x} = \dfrac{1}{{\log (1999!)}}[\log (1 \times 2 \times 3....\operatorname{l} 998 \times 1999)] $
Rewrite the above equation – applying the multiplicative property where $ a \times b = b \times a $
 $ \sum\limits_{x = 1}^{1999} {{{\log }_n}x} = \dfrac{1}{{\log (1999!)}}[\log (1999 \times \operatorname{l} 998.....3 \times 2 \times 1)] $
Apply the concept of the factorial of the number and simplify the above equation – where $ n! = n \times (n - 1) \times (n - 2).....1 $
 $ \sum\limits_{x = 1}^{1999} {{{\log }_n}x} = \dfrac{1}{{\log (1999!)}}[\log (1999!)] $
The same numerator and the denominator cancel each other in the above equation –
\[\sum\limits_{x = 1}^{1999} {{{\log }_n}x} = 1\]
So, the correct answer is “Option A”.

Note: Remember all the nine basic rules of the logarithms to solve these types of problems accurately and efficiently. Do simplification carefully. Also, remember the concept of factorial and apply wisely.
The factorial of the positive integer ”n” is expressed as the $ n! $ . Also remember that the $ 0! = 1 $ . The factorial of the number can be obtained by the number multiplied by the “number minus one” and then multiplied with the “number minus two” and so one till one. Factorial is only for the positive natural numbers.