Question

In the below logarithmic series if $n=2017!$, then what is the value of the below series:
$\dfrac{1}{{{\log }_{2}}n}+\dfrac{1}{{{\log }_{3}}n}+\dfrac{1}{{{\log }_{4}}n}+.........+\dfrac{1}{{{\log }_{2017}}n}$
(a)0
(b)1
(c) n/2
(d) n

Answer 1

Hint: We have to evaluate the given logarithmic series which we can write by using the property of a log that $\dfrac{1}{{{\log }_{b}}a}={{\log }_{a}}b$. Now, the series will look like ${{\log }_{n}}2+{{\log }_{n}}3+....+{{\log }_{n}}2017$. In this new series, we are going to apply the property of logarithm that ${{\log }_{a}}x+{{\log }_{a}}y={{\log }_{a}}\left( xy \right)$ and then solve the logarithmic series.

Complete step-by-step answer:
The logarithmic series given in the above question is:
$\dfrac{1}{{{\log }_{2}}n}+\dfrac{1}{{{\log }_{3}}n}+\dfrac{1}{{{\log }_{4}}n}+.........+\dfrac{1}{{{\log }_{2017}}n}$
To simplify this above series we are going to use the property of the logarithm that:
 $\dfrac{1}{{{\log }_{b}}a}={{\log }_{a}}b$
Using this property in the logarithmic series we get,
${{\log }_{n}}2+{{\log }_{n}}3+{{\log }_{n}}4+...........+{{\log }_{n}}2017$
As you can see in the above series that the base of the log is same i.e. “n” so we can use the property of logarithm that:
${{\log }_{a}}x+{{\log }_{a}}y={{\log }_{a}}\left( xy \right)$
We have shown the formula for two logarithms but we can extend this formula for more than two logarithms also.
${{\log }_{n}}\left( 2.3.4.5......2017 \right)$
We can also multiply 1 in the argument of the above logarithm.
${{\log }_{n}}\left( 1.2.3.4.5......2017 \right)$……….. Eq. (1)
We know that:
$n!=n\left( n-1 \right)\left( n-2 \right).....1$
So we can write 2017! as:
$1.2.3.4.5.......2017=2017!$
So, we can use this factorial notation in the argument of the logarithm given in the eq. (1).
${{\log }_{n}}2017!$
It is given that $n=2017!$ so substituting this value in the above logarithmic expression we get,
${{\log }_{2017!}}\left( 2017! \right)$
And we know that when the base and argument of the logarithm is the same then the value of that logarithm is 1 so as you can see in the above expression that base and argument is the same which is equal to 2017! So the answer of the above logarithm is 1.
 ${{\log }_{2017!}}\left( 2017! \right)=1$
Hence, the value of the given logarithmic series is 1.
Hence, the correct option is (b).

Note: You might think of how we know which property of the logarithm will be used where. For instance, first of all we have used the property $\dfrac{1}{{{\log }_{b}}a}={{\log }_{a}}b$. We have used this property because it will make the base same in all the logarithms and we can then easily multiply the argument of the logarithms and multiplication of all the arguments will give us 2017!. Now, you might ask why we multiply all the arguments of logarithm because the base of the logarithm is n which is given as 2017! And multiplication of the arguments will give us 2017! Which makes the base and argument similar and we can easily solve the problem.