Question

What is the formula of \[\log \left( {\dfrac{1}{x}} \right)\] ?

Answer 1

Hint: Here in the given question, we have to find the formula of given logarithm function. This can be found, by using a quotient rule of logarithmic function i.e., \[\log \left( {\dfrac{m}{n}} \right) = \log m - \log n\] . Otherwise also find by using a rule of negative rule of exponent i.e., \[{a^{ - x}} = \dfrac{1}{{{a^x}}}\] and by further simplification we get the required solution.

Complete step-by-step answer:
The function from positive real numbers to real numbers to real numbers is defined as \[{\log _b}:{R^ + } \to R \Rightarrow {\log _b}\left( x \right) = y\] , if \[{b^y} = x\] , is called logarithmic function or the logarithm function is the inverse form of exponential function.
There are some basic logarithms properties
1. Product rule :- \[\log \left( {mn} \right) = \log m + \log n\]
2. Quotient rule :- \[\log \left( {\dfrac{m}{n}} \right) = \log m - \log n\]
3. Power rule :- \[\log \left( {{m^n}} \right) = n.\log m\]
Consider the given logarithm function:
 \[ \Rightarrow \log \left( {\dfrac{1}{x}} \right)\]
Applying a quotient rule of logarithm function, we have
 \[ \Rightarrow \log \left( 1 \right) - \log \left( x \right)\]
As by the standard logarithm calculator or logarithm table the value of \[\log \left( 1 \right) = 0\] , then
 \[ \Rightarrow 0 - \log \left( x \right)\]
 \[ \Rightarrow - \log \left( x \right)\]
Therefore, the formula \[\log \left( {\dfrac{1}{x}} \right) = - \log \left( x \right)\] .
Otherwise
It can also solve by using exponential rules:
Exponential notation is an alternative method of expressing numbers. Exponential numbers take the form an, where a is multiplied by itself n times. In exponential notation, a is termed the base while n is termed the power or exponent or index.
Some exponent rules are:
Product rule :- \[{a^x} + {a^y} = {a^{x + y}}\]
Quotient rule :- \[\dfrac{{{a^x}}}{{{a^y}}} = {a^{x - y}}\]
Power rule :- \[{\left( {{a^x}} \right)^y} = {a^{xy}}\]
Negative rule :- \[{a^{ - x}} = \dfrac{1}{{{a^x}}}\]
Zero rule :- \[{a^0} = 1\]
Now, consider the given function
 \[ \Rightarrow \log \left( {\dfrac{1}{x}} \right)\]
Applying a negative rule of exponential, we have
 \[ \Rightarrow \log \left( {{x^{ - 1}}} \right)\]
By the power rule of logarithm function, then
 \[ \Rightarrow \left( { - 1} \right)\log \left( x \right)\]
 \[ \Rightarrow - \log \left( x \right)\]
Hence, by both the methods we get the formula \[\log \left( {\dfrac{1}{x}} \right) = - \log \left( x \right)\]

Note: The question contains the log terms we must know the logarithmic properties which are the standard properties. By applying properties we can solve the question in an easy manner. We apply the formula \[{\log _b}\left( x \right) = y\] that can be written as \[{b^y} = x\] . where it is necessary. Hence, we obtain the desired result.