Question

# Find the value of the following expression-$\dfrac{1}{{{\log }_{2}}N}+\dfrac{1}{{{\log }_{3}}N}+\dfrac{1}{{{\log }_{4}}N}+....\dfrac{1}{{{\log }_{100}}N}$=? (for N $\ne$1)(a) $\dfrac{1}{{{\log }_{100!}}N}$(b) $\dfrac{1}{{{\log }_{99!}}N}$(c) $\dfrac{99}{{{\log }_{100!}}N}$(d) $\dfrac{99}{{{\log }_{99!}}N}$

Hint: To solve this problem, we need to be aware about the basic logarithmic properties. These are-
${{\log }_{x}}y=\dfrac{1}{{{\log }_{y}}x}$ -- (1)
${{\log }_{x}}a+{{\log }_{x}}b={{\log }_{x}}ab$ -- (2)
Thus, to solve this problem, we would first use property (1) and then use property (2) to get the correct answer. In addition, we would also use the formula of factorial of a number-
$N!=1\times 2\times 3\times ....\times N$

We can see that the options are a single term of logarithm. Thus, to do so, we would somehow have to use property (2), that is, ${{\log }_{x}}a+{{\log }_{x}}b={{\log }_{x}}ab$to combine the terms. However, currently all terms are in the form of reciprocal of logarithms, thus we cannot use property (2). Thus, we would first start by using property (1), that is, ${{\log }_{x}}y=\dfrac{1}{{{\log }_{y}}x}$ on all the terms of the problem. Doing so, we have,
=${{\log }_{N}}2+{{\log }_{N}}3+{{\log }_{N}}4+....{{\log }_{N}}100$
Now, by doing so, we can see that all the terms have the same logarithmic base (that is N). Thus, we can use the extension of property (2), this is –
${{\log }_{x}}a+{{\log }_{x}}b+{{\log }_{x}}c+....+{{\log }_{x}}z={{\log }_{x}}abcd...z$ -- (3)
Thus, we can use property (3) to solve,
=${{\log }_{N}}(2\times 3\times 4\times ...\times 100)$
Now, we use the definition of factorial of a whole number, that is,
$N!=1\times 2\times 3\times ....\times N$
Also, 0! = 1
Thus, now we have,
Since, $2\times 3\times 4\times ...\times 100=1\times 2\times 3\times 4\times ...\times 100$=100!
=${{\log }_{N}}(100!)$
Now, since, all the options are in reciprocal, we simply use property (1) to get the answer,
=$\dfrac{1}{{{\log }_{100!}}N}$
Hence, the correct option is (a).

Note: To solve problems related to logarithms, it is always a good idea to be aware about all the various properties related to manipulation of the bases and otherwise. In this problem, we used the property ${{\log }_{x}}a+{{\log }_{x}}b={{\log }_{x}}ab$. However, one must also be aware of the property, ${{\log }_{x}}a-{{\log }_{x}}b={{\log }_{x}}\dfrac{a}{b}$.