More about Logarithms: There are certain processes in nature where the growth/decay of some quantity has very peculiar properties, for example, the rate of growth of some bacteria, the rate of decay of radioactive materials, etc. These processes own a typical characteristic which is known as exponential growth/decay, and the functions representing them are called exponential functions. Whereas, the logarithmic scales reduce wide-ranging quantities into small scopes. For example, the decibel \(\left( {{\text{dB}}} \right)\) is a unit which used to express the ratio as logarithms for signal power and amplitude. In chemistry, the potential of hydrogen \(\left( {{\text{pH}}} \right)\) is a logarithmic measure for the acidity of an aqueous solution.

What are Logarithm Functions?

Every positive real number \(N\) can be expressed in exponential form as \(N = {a^x}\)
where \(a\) is a positive real number other than unity and is called the base, and \(x\) is called the exponent.
We can write the above relation in logarithmic form as
\({\log _a}N = x\)
Hence, the two relations \({a^x} = N\) and \({\log _a}N = x\) are identical where \(N > 0,a > 1\)
Note: An exponential function is defined by the formula \(f\left( x \right) = {a^x},\) where \(a\) is a real number and is known as ‘base’ and \(x\) is called exponent, \(x \in R,a > 0\) but \(a \ne 1.\) Hence, the domain of \(f\left( x \right) = {a^x}\) is a set of real numbers and the range of \(f\left( x \right) = {a^x}\) is \(\left( {0,\infty } \right).\)

Definition of Logarithm

Consider \({a^x} = N,\) here we are expressing \(N\) in terms of \(a\) and \(x.\) Another way to write the same relation is \(a = {N^{\frac{1}{x}}}\) Here we are expressing \(a\) in terms of \(N\) and \(x.\) If we have to express \(x\) in terms of the other two parameters \(a\) and \(N\) then we have to use logarithms as, \(x = {\log _a}N.\)
The logarithm can be defined as, “The logarithm \(\left( x \right)\) of a positive number \(\left( N \right)\) to a positive base \(\left( a \right)\) is the index to which the base is raised in order to equal the given number”.
\({a^x} = N\)
\( \Rightarrow x = {\log _a}N\)
Note:
1. In the equation, \(N\) is a positive number, and \(a\) is any positive number other than \(1.\)
2. Unity cannot be the base of the logarithm, as \({\log _1}N\) is not computable. Also,\({\log _1}1\) will have infinitely many solutions and will not be unique, which is necessary for the functional notation.

Types of Logarithms

Logarithms can be expressed to any base (from definition positive number other than \(1\)).
Logarithms from one base can be converted into logarithms to any other base. However, there are two types of logarithms based on commonly used bases.
i. Common Logarithms
These are logarithms expressed to the base \(10.\) In most of the problems under logarithms, we deal with the common logarithms. If logarithms are mentioned without mentioning a specific base, it can be taken to the logarithms of base \(10.\) If \(a = 10,\) then we write \(\log b\) instead of \({\log _{10}}b.\)
ii. Natural Logarithms or Napierian Logarithms
These are logarithms expressed to the base of a constant \(e.\) If \(a = e,\) then we write \(\ln N\) instead of \({\log _e}N.\) If the base \(a\) is Euler number \(e,\) then the exponential function \({e^x}\) is known as the natural exponential function, where the value of \(e\) is approximately \(2.71828.\)

Conversion of Bases

We can write
\({\log _e}a = {\log _{10}}a{\log _e}10\)
\({\log _{10}}a = \frac{{{{\log }_e}a}}{{{{\log }_e}10}}\)
\({\log _{10}}a = 0.434\,{\log _e}a\)
This transformation is used to convert natural logarithm to common logarithm.
Note:
1. The logarithm of a number is unique, i.e., no number can have two different logarithms to a given base.
2. Domain of function \(y = {\log _a}x\) is \(\left( {0,\infty } \right)\) and the range \(\left( { – \infty ,\infty } \right).\)
3. When \(x \to 0,\) then \({\log _a}x \to – \infty \) (for \(a > 1\) and \({\log _a}x \to \infty \) (for \(0 < a < 1\)) because \(y = {\log _a}x \Rightarrow x = {a^y}\) which approaches zero if \(y \to – \infty \) as \({a^{ – \infty }} = 0\forall a > 1.\)
4. When \(a \in \left( {0,1} \right),x = {a^y}\) approaches to zero if \(y \to \infty \) as \({a^\infty } = 0\forall 0 < a < 1.\)

Properties of Logarithm

Let \(a\) and \(b\) be positive numbers such that \(a > 1,b > 1,\) then

Name of the property	Definition	Description
Zero Rule	\({\log _a}1 = 0\)	Because \(0\) is the power to which \(a\) must be raised to obtain \(1\).
Identity Rule	\({\log _a}a = 1\)	Since \(1\) is the power to which \(a\) must be raised to obtain \(a\).
Exponent of Log Rule and Log of Exponent Rule	1. \({a^{{{\log }_a}N}} = N\) 2. \({\log _a}{a^N} = N\)	As \(N\) is the power to which \(a\) must be raised to obtain \({a^N}.\)
Product Rule	\({\log _m}\left( {a.b} \right) = {\log _m}\left| a \right| + {\log _m}\left| b \right|;\left( {a.b} \right) > 0\)	The logarithm of the product of two numbers to a certain base is the sum of the logarithms of the numbers to that base.
Quotient Rule	\({\log _m}\left( {\frac{a}{b}} \right) = {\log _m}\left| a \right| – {\log _m}\left| b \right|;\left( {a.b} \right) > 0\)	The logarithm of the quotient of two numbers of the same signs is equal to the difference of their magnitude’s logarithms, the base remaining the same throughout.
Power Rule	\({\log _a}{N^k} = k{\log _a}\left| N \right|\) (\(\forall k \in R\) for \(N > 0\)) or (\(k\) is such that \({N^k} > 0\) if \(N < 0\)).	The logarithm of the permissible (power of a number) is equal to the product of the power and the logarithm of the magnitude of the number (base remaining the same).
	\({\log _{{a^k}}}N = \left( {\frac{1}{k}} \right) \cdot {\log _{\left| a \right|}}N;{a^k} > 0\)
	\({\log _b}a = {\log _c}a \cdot {\log _b}c\)
	\({\log _b}a = \frac{{\log a}}{{\log b}}\)
	\({\log _b}a \cdot {\log _a}b = 1\)
	\({a^{{{\log }_m}b}} = {b^{{{\log }_m}a}}\)

Logarithmic Inequation

To solve inequations involving logarithmic functions, we need to understand the nature of the graph of \(y = {\log _a}x.\)

Graph	Observations and Conclusions
\(y = {\log _a}x,a > 1\)	1. Domain of the function \(\left( {0,\infty } \right).\) 2. Range of the function \(\left( { – \infty ,\infty } \right).\) 3. Function is strictly increasing 4. \(y = {\log _a}x\) is negative if \(x \in \left( {0,1} \right).\) 5. \(y = {\log _a}x\) is zero if \(x = 1.\) 6. \(y = {\log _a}x\) is positive if \(x \in \left( {1,\infty } \right).\) 7. If \({\log _a}p > {\log _a}q\) then \(p > q\)
\(y = {\log _a}x,0 < a < 1\)
	1. Domain of the function \(\left( {0,\infty } \right)\) 2. Range of the function \(\left( { – \infty ,\infty } \right).\) 3. Function is strictly decreasing 4. \(y = {\log _a}x\) is negative if \(x \in \left( {1,\infty } \right).\) 5.\(y = {\log _a}x\) is zero if \(x = 1.\) 6. \(y = {\log _a}x\) is positive if \(x \in \left( {0,1} \right).\) 7. If \({\log _a}p > {\log _a}q\) then \(p < q.\)

Now, we can discuss the logarithm inequality

Base	Logarithmic Inequality	Exponential Inequality
\(y > 1\)	\({\log _y}x > z\)	\(x > {y^z}\)
\(y > 1\)	\({\log _y}x < z\)	\(x < {y^z}\)
\(0 < y < 1\)	\({\log _y}x > z\)	\(0 < x < {y^z}\)
\(0 < y < 1\)	\({\log _y}x < z\)	\(x > {y^z}\)
\(a > 1\)	\({\log _a}p > {\log _a}q\)	\(p > q\)
\(a < 1\)	\({\log _a}p < {\log _a}q\)	\(p < q\)

Differentiating Logarithms

Let \(y = {\log _a}x…\left( 1 \right)\)
Here, \(a > 1\)
Let \(\delta x\) be a small increment in \(x\) and \(\delta y\) the corresponding increment in \(y\)
\(\therefore y + \delta y = {\log _a}\left( {x + \delta x} \right)…..\left( 2 \right)\)
Subtracting \(\left( 1 \right)\) from \(\left( 2 \right),\) we have
\(\delta y = {\log _a}\left( {x + \delta x} \right) – {\log _a}x\)
\(\delta y = {\log _a}\left( {\frac{{x + \delta x}}{x}} \right) = {\log _a}\left( {1 + \frac{{\delta x}}{x}} \right)…\left( 3 \right)\)
Dividing both sides of eq. \(\left( 3 \right)\) by \(\delta x,\) we have
\(\frac{{\delta y}}{{\delta x}} = \frac{1}{{\delta x}}{\log _a}\left( {1 + \frac{{\delta x}}{x}} \right)\)
\(\frac{{\delta y}}{{\delta x}} = \frac{1}{x}.\frac{x}{{\delta x}}{\log _a}\left( {1 + \frac{{\delta x}}{x}} \right)\)
\(\frac{{\delta y}}{{\delta x}} = \frac{1}{x}{\log _a}{\left( {1 + \frac{{\delta x}}{x}} \right)^{\frac{x}{{\delta x}}}}\)
Proceeding to limits as \(\delta x \to 0\), we have
\(\mathop {\lim }\limits_{\delta x \to 0} \frac{{\delta y}}{{\delta x}} = \mathop {\lim }\limits_{\delta x \to 0} \left[ {\frac{1}{x}{{\log }_a}{{\left( {1 + \frac{{\delta x}}{x}} \right)}^{\frac{x}{{\delta x}}}}} \right]\)
\( \Rightarrow \frac{{dy}}{{dx}} = \frac{1}{x}{\log _a}e\) \(\left[ {\because \mathop {\lim }\limits_{\delta x \to 0} {{\left( {1 + \frac{{\delta x}}{x}} \right)}^{\frac{x}{{\delta x}}}} = e} \right]\)
Hence, \(\frac{d}{{dx}}{\log _a}x = \frac{1}{x}{\log _a}e\)
Note:
1. \(\frac{d}{{dx}}{\log _e}x = \frac{1}{x}\)
2. \(\frac{d}{{dx}}{a^x},x > 0 = {a^x}{\log _e}a\)

Solved Examples on  Logarithm

Q.1. Express \({625^{\frac{4}{9}}} = 25x\) in the form \(\log \) of \(x\) to the base \(5.\)
Ans:
Given: \({625^{\frac{4}{9}}} = 25x\)
\( \Rightarrow \frac{{{5^{\frac{{16}}{9}}}}}{{25}} = x\)
\( \Rightarrow {5^{\frac{{16}}{9} – 2}} = x\)
\( \Rightarrow {5^{ – \frac{2}{9}}} = x\)
\( \Rightarrow {\log _5}x = – \frac{2}{9}\)
Hence, \({\log _5}x = – \frac{2}{9}.\)

Q.2. Find the logarithm of \(32\sqrt[5]{4}\) to base \(2\sqrt 2 .\)
Ans:
Let \(N\) be the required logarithm;
\(\therefore {\log _{2\sqrt 2 }}32\sqrt[5]{4} = N\)
By definition, \({\left( {2\sqrt 2 } \right)^N} = 32\sqrt[5]{4}\)
\(\therefore {\left( {{{2.2}^{\left( {\frac{1}{2}} \right)}}} \right)^N} = {2^5} \cdot {2^{\frac{2}{5}}} \Rightarrow {2^{\frac{3}{2}N}} = {2^{5 + \left( {\frac{2}{5}} \right)}}\)
\( \Rightarrow \frac{3}{2}N = \frac{{27}}{5} \Rightarrow N = \frac{{18}}{5}\)
So, the value of \(N\) is \(\frac{{18}}{5}\)
Hence, the required logarithm is \(\frac{{18}}{5}\)

Q.3. Solve \({\log _{ab}}\frac{{\sqrt[3]{a}}}{{\sqrt b }}\) if \({\log _{ab}}a = 4.\)
Ans: \({\log _{ab}}\frac{{\sqrt[3]{a}}}{{\sqrt b }} = {\log _{ab}}\sqrt[3]{a} – {\log _{ab}}\sqrt b \)
\({\log _{ab}}\frac{{\sqrt[3]{a}}}{{\sqrt b }} = \frac{1}{3}{\log _{ab}}a – \frac{1}{2}{\log _{ab}}b\)
\({\log _{ab}}\frac{{\sqrt[3]{a}}}{{\sqrt b }} = \frac{4}{3} – \frac{1}{2}{\log _{ab}}b\)
It remains to find the quantity \({\log _{ab}}b.\)
Since \(1 = {\log _{ab}}ab = {\log _{ab}}a + {\log _{ab}}b = 4 + {\log _{ab}}b\)
\( \Rightarrow 1 = 4 + {\log _{ab}}b \Rightarrow {\log _{ab}}b = – 3\)
\(\therefore {\log _{ab}}\frac{{\sqrt[3]{a}}}{{\sqrt b }} = \frac{4}{3} – \frac{1}{2} \times \left( { – 3} \right) = \frac{{17}}{6}\)

Q.4. Solve the equation \({\log _3}\left( {23 + 4{{\log }_3}\left( {2x – 3} \right)} \right) = 3.\)
Ans: \({\log _3}\left( {23 + 4{{\log }_3}\left( {2x – 3} \right)} \right) = 3\)
\(\therefore 23 + 4{\log _3}\left( {2x – 3} \right) = {3^3}\)
\( \Rightarrow {\log _3}\left( {2x – 3} \right) = 1\)
\( \Rightarrow 2x – 3 = 3\)
\( \Rightarrow 2x = 6\)
\(\therefore x = 3\)

Q.5. Find the derivative of the \(f\left( x \right) = \ln x \cdot {\log _4}x.\)
Ans: Given: \(f\left( x \right) = \ln x \cdot {\log _4}x.\)
Now, \(\frac{d}{{dx}}\left( {f\left( x \right)} \right) = \frac{d}{{dx}}\left( {\ln x} \right) \cdot {\log _4}x + \ln x\frac{d}{{dx}}{\log _4}x\)
\(f’\left( x \right) = \frac{{{{\log }_4}x}}{x} + \frac{{\ln x}}{{x\ln 4}}\) \(\left[ {\because \frac{d}{{dx}}{{\log }_e}x = \frac{1}{x},\frac{d}{{dx}}{{\log }_a}x = \frac{1}{x}{{\log }_a}e = \frac{1}{{x\left( {\ln a} \right)}}} \right]\)
\(\therefore \frac{d}{{dx}}\left( {f\left( x \right)} \right) = \frac{{{{\log }_4}x}}{x} + \frac{{\ln x}}{{x\ln 4}}\)

Summary

The logarithm can be defined as “The logarithm of any number (say a positive number \(N\)) to a given base (say any positive number other than \(1\)) is the index or the power (say \(x\)) to which the base must be raised in order to equal the given number” i.e., \({a^x} = N \Rightarrow {\text{x}} = {\log _a}N.\) There are two types of logarithms common logarithms with base \(10\) and natural logarithms or Napierian logarithms with base \(e.\) We can convert the base of logarithm by using the formula \({\log _{10}}a = 0.434{\log _e}a.\) The logarithm of a number is unique, i.e. no number can have two different logarithms to a given base. Domain of function \(y = {\log _a}x\) is \(\left( {0,\infty } \right)\) and the range \(\left( { – \infty ,\infty } \right).\)

Frequently Asked Questions (FAQs)

The students must be having questions regarding logarithms. Here we have compiled a few to clear all the doubts.

Q.1. What is the power rule of logarithms?
Ans: When the logarithmic term has an exponent, then by logarithm power rule we can transfer the exponent to the front of the logarithm. Along with the product rule and the quotient rule, the logarithm power rule can be used for expanding and condensing logarithms. This is given by \({\log _a}{N^k} = k{\log _a}\left| N \right|\)

Q.2. How many types of logarithms are there?
Ans: There are only two types of logarithms common logarithms with base \(10\) and natural logarithms or Napierian logarithms with base \(e,\) where \(e \approx 2.71828.\)

Q.3. What do you mean by logarithm?
Ans: The logarithm can be defined as, “The logarithm of any positive number to a given (positive) base other than \(1\) is the index or the power \(x,\) to which the base must be raised to equal the given number”.
\({a^x} = N\)
\( \Rightarrow x = {\log _a}N\)

Q.4. How are logarithms used in real life?
Ans: The logarithmic scales reduce wide-ranging quantities into small scopes.
There are many uses of logarithm in real-life few of them are listed below:
i. The decibel \(\left( {{\text{dB}}} \right)\) is a unit used to express the ratio as logarithms, primarily for signal power and amplitude.
ii. In chemistry, the potential of hydrogen \(\left( {{\text{pH}}} \right)\) is a logarithmic measure for the acidity of an aqueous solution.

Q.5. What are the \(7\) rules of logarithms?
Ans: Rules of Logarithms are as follows:
1. Product Rule: \({\log _m}\left( {a.b} \right) = {\log _m}\left| a \right| + {\log _m}\left| b \right|;a.b > 0\)
2. Quotient Rule: \({\log _m}\left( {\frac{a}{b}} \right) = {\log _m}\left| a \right| – {\log _m}\left| b \right|;a.b > 0\)
3. Power Rule: \({\log _a}{N^k} = k{\log _a}\left| N \right|\) (\(\forall k \in R\) for \(N > 0\) or \(k\) is such that \({N^k} > 0\) if \(N < 0\)).
4. Zero Rule: \({\log _a}1 = 0\)
5. Identity Rule: \({\log _a}a = 1\)
6. Log of Exponent Rule: \({\log _a}{a^N} = N\)
7. Exponent of Log Rule: \({a^{{{\log }_a}N}} = N\)

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