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

Last updated date: 23rd Mar 2023

•

Total views: 306k

•

Views today: 4.88k

Answer

Verified

306k+ 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, ${{\log }_{x}}a-{{\log }_{x}}b={{\log }_{x}}\dfrac{a}{b}$.

Recently Updated Pages

If ab and c are unit vectors then left ab2 right+bc2+ca2 class 12 maths JEE_Main

A rod AB of length 4 units moves horizontally when class 11 maths JEE_Main

Evaluate the value of intlimits0pi cos 3xdx A 0 B 1 class 12 maths JEE_Main

Which of the following is correct 1 nleft S cup T right class 10 maths JEE_Main

What is the area of the triangle with vertices Aleft class 11 maths JEE_Main

KCN reacts readily to give a cyanide with A Ethyl alcohol class 12 chemistry JEE_Main

Trending doubts

What was the capital of Kanishka A Mathura B Purushapura class 7 social studies CBSE

Difference Between Plant Cell and Animal Cell

Write an application to the principal requesting five class 10 english CBSE

Ray optics is valid when characteristic dimensions class 12 physics CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Tropic of Cancer passes through how many states? Name them.

Write the 6 fundamental rights of India and explain in detail

Write a letter to the principal requesting him to grant class 10 english CBSE

Name the Largest and the Smallest Cell in the Human Body ?