# 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) $\dfrac{99}{{{\log }_{99!}}N}$

Answer

Verified

362.1k+ views

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-

$N!=1\times 2\times 3\times ....\times N$

Complete step-by-step answer:

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, ${{\log }_{x}}a-{{\log }_{x}}b={{\log }_{x}}\dfrac{a}{b}$.

${{\log }_{x}}y=\dfrac{1}{{{\log }_{y}}x}$ -- (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-

$N!=1\times 2\times 3\times ....\times N$

Complete step-by-step answer:

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, ${{\log }_{x}}a-{{\log }_{x}}b={{\log }_{x}}\dfrac{a}{b}$.

Last updated date: 26th Sep 2023

•

Total views: 362.1k

•

Views today: 6.62k

Recently Updated Pages

What is the Full Form of DNA and RNA

What are the Difference Between Acute and Chronic Disease

Difference Between Communicable and Non-Communicable

What is Nutrition Explain Diff Type of Nutrition ?

What is the Function of Digestive Enzymes

What is the Full Form of 1.DPT 2.DDT 3.BCG

Trending doubts

How do you solve x2 11x + 28 0 using the quadratic class 10 maths CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Summary of the poem Where the Mind is Without Fear class 8 english CBSE

Difference Between Plant Cell and Animal Cell

What is the basic unit of classification class 11 biology CBSE

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

One cusec is equal to how many liters class 8 maths CBSE

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

Give 10 examples for herbs , shrubs , climbers , creepers