Question

What is $$ $\dfrac{1}{{{{\log }_2}N}} + \dfrac{1}{{{{\log }_3}N}} + \dfrac{1}{{{{\log }_4}N}}... + \dfrac{1}{{{{\log }_{100}}N}}$ equal to $(N \ne 1)$ ?

Answer 1

Hint : In order to determine the given logarithmic function by using the base rule formula,
Change of base rule: ${\log _b}m = \dfrac{{{{\log }_a}m}}{{{{\log }_a}b}}$
Base switch rule: ${\log _b}(a) = \dfrac{1}{{{{\log }_a}(b)}}$
We use the logarithmic formula is $\log 2 + \log 3 + \log 4 + ..... + \log N = \log (N! - 1)$ then get the required solution.

Complete step-by-step answer :
A logarithm is known as the power to which a number must be raised in order to obtain any other values. It is the simplest way to express large numbers. A logarithm has many essential properties that demonstrate that logarithm multiplication and division can also be written in the form of logarithm addition and subtraction.
In this problem,
We are given the logarithm function, $\dfrac{1}{{{{\log }_2}N}} + \dfrac{1}{{{{\log }_3}N}} + \dfrac{1}{{{{\log }_4}N}}... + \dfrac{1}{{{{\log }_{100}}N}}$
Let the given function as ‘A’:
$A = \dfrac{1}{{{{\log }_2}N}} + \dfrac{1}{{{{\log }_3}N}} + \dfrac{1}{{{{\log }_4}N}}... + \dfrac{1}{{{{\log }_{100}}N}}$ ------------(1)
On comparing the equation (1) with the basic switch rule, ${\log _b}(a) = \dfrac{1}{{{{\log }_a}(b)}}$ , then
Now, we have to change the equation as per the rule, then
${\rm A} = {\log _N}(2) + {\log _N}(3) + {\log _N}(4)... + {\log _N}(100)$
Now, changing the base rule,${\log _b}m = \dfrac{{{{\log }_a}m}}{{{{\log }_a}b}}$
$= \dfrac{{\log 2}}{{\log N}} + \dfrac{{\log 3}}{{\log N}} + \dfrac{{\log 4}}{{\log N}}... + \dfrac{{\log 100}}{{\log N}}$
Taking out $\dfrac{1}{{\log N}}$ from the bracket, we get
$= \dfrac{1}{{\log N}}(\log 2 + \log 3 + \log 4 + .... + \log 100)$
We use the logarithmic formula is $\log 2 + \log 3 + \log 4 + ..... + \log N = \log (N! - 1)$
$= \dfrac{1}{{\log N}}\log (100! - 1)$
$= \dfrac{{\log (100! - 1)}}{{\log N}}$
Now, compare the change of basic rule formula is ${\log _b}m = \dfrac{{{{\log }_a}m}}{{{{\log }_a}b}}$
${\rm A} = {\log _N}(100! - 1)$
Substitute the value in equation (1), then
$\Rightarrow \dfrac{1}{{{{\log }_2}N}} + \dfrac{1}{{{{\log }_3}N}} + \dfrac{1}{{{{\log }_4}N}}... + \dfrac{1}{{{{\log }_{100}}N}} = {\log _N}(100! - 1)$
Hence, $\dfrac{1}{{{{\log }_2}N}} + \dfrac{1}{{{{\log }_3}N}} + \dfrac{1}{{{{\log }_4}N}}... + \dfrac{1}{{{{\log }_{100}}N}}$ $= {\log _N}(100! - 1)$ . Since, $(N \ne 1)$ .

Note : There are certain rules based on which logarithmic operations can be performed. The names of these rules are:
Power rule or Exponential Rule: In the exponential rule, the logarithm of m with a rational exponent is equal to the exponent times its logarithm.
Division rule: The division of two logarithmic values is equal to the difference of each logarithm. Derivative of log: If $f(x) = {\log _{10}}(x)$ then the derivative of $f(x)$ is given by;

$$f'\left( x \right){\text{ }} = {\text{ }}1/\left( {x{\text{ }}ln\left( b \right)} \right) \\\ f'(x) = \dfrac{1}{{(x\ln (b))}} \;$$

Integral of log: $\int {{{\log }_b}(x)dx = x({{\log }_b}(x) - \dfrac{1}{{\ln (b)}}) + C}$
Some other properties of logarithmic functions are mentioned below:
${\log _b}b = 1 \\\ {\log _b}1 = 0 \\\ {\log _b}0 = undefined \;$
Product rule: In this rule, the multiplication of two logarithmic values is equal to the addition of their individual logarithms.