**Logarithms** (and **Antilogarithms**) are primarily used to make calculations of ** powers and roots** of any number easier, although they have numerous other real world applications.

Logarithm of a number ** N** with respect another number

Therefore by definition of word logarithm we have

\(\Rightarrow\)

Where

**log**(_{B}1=0*That is*)**log of 1 to any base is always 0****log**(_{B}B=1*That is*)**log of a number with same base is equal to 1**

**Base 10:**Such logarithms are known as**Common Logarithms.****Base 'e':**Such logarthims are known as**Natural Logarithms.****Base 2:**Such logarithms are known as**Binary Logarithms.**

There are 4 primary rules that apply to logarithms irrespective of the base used to calculate the logarithm.

**Production Addition Rule:**This rule states that the logarithm of product of any numbers is equal to the sum of logarithms of those numbers. That is,

**log**(_{k}**A x B x C x D... x**) =*N***log**_{k}A + log_{k}B + log_{k}C + log_{k}D +... log_{k}N

We will prove the rule for the following simple case:*Proof:*

**log**_{k}(A x B) = log_{k}A + log_{k}B

Let**log**_{k}A = T

\(\Rightarrow\)**A=k**^{T}

Now, Let**log**_{k}B = S

\(\Rightarrow\)**B =k**^{S}

Therefore

**A x B = k**^{T}x k^{S}

\(\Rightarrow\)**A x B = k**^{T+S}

\(\Rightarrow\)**log**(By Definition of Logarithm)_{k}(A x B) = T+S

\(\Rightarrow\)**log**_{k}(A x B) = log_{k}A + log_{k}B

An interesting rule that can be derived through this rule itself is the following

**log**_{k}A^{N }= N log_{k}A

*Proof:*

**log**_{k}A^{N }= log_{k}(A x A x A x....*N Times***A) = log**_{k}A + log_{k}A + log_{k}A +....*N Times***log**=_{k}A**N x log**_{k}A

**Division Subtraction Rule:**This rule states that the logarithm of result of division two of any two numbers is equal to the difference of logarithms of the dividend and the divisor. That is,

**log**_{k }(A/B) = log_{k}A - log_{k}B

*Proof:*

Let**log**_{k}A = T

\(\Rightarrow\)**A=k**^{T}

Now, Let**log**_{k}B = S

\(\Rightarrow\)**B =k**^{S}

Therefore

**A/B = k**^{T}/ k^{S}

\(\Rightarrow\)**A/B = k**^{T-S}

\(\Rightarrow\)**log**(By Definition of Logarithm)_{k}(A/B) = T-S

\(\Rightarrow\)**log**_{k}(A/B) = log_{k}A - log_{k}B

**Reciprocal Rule:**This rule states that the logarithm of number N to the base B is reciprocal of logarithm of number B to the base N. That is,

**log**_{B}N = 1 / log_{N}B

*Proof:*

Let**log**_{B }N = P

\(\Rightarrow\)**B**^{P}= N

Taking log to the base N on both sides

**log**_{N}B^{P}= log_{N}N

\(\Rightarrow\)**P x log**_{N }B = 1

\(\Rightarrow\)**log**_{N}B = 1/P

\(\Rightarrow\)**log**_{N}B = 1 / log_{B}N

**Chain Rule:**This rule can be best stated with the following formula,

**log**_{A}B x log_{B}C = log_{A}C

*Proof:*

Let**log**_{A}B = T

\(\Rightarrow\)**B = A**^{T}

Now Let**log**_{B}C = N

\(\Rightarrow\)**C = B**^{N}

\(\Rightarrow\)**C = (A**^{T})^{ N}

Taking log to the base A on both sides

**log**_{A}C = log_{A}(A^{T})^{ N}

\(\Rightarrow\)**log**_{A}C = log_{A}(A)^{TN}

\(\Rightarrow\)**log**_{A}C = T x N x log_{A}(A)

\(\Rightarrow\)**log**_{A}C = T x N x 1

\(\Rightarrow\)**log**_{A}C = log_{A}B x log_{B}C

\(\Rightarrow\)

\(\Rightarrow\)

Since both

** Anti-Logarithms** provide the actual number whose logarithmic value for a particular base is given.

Because of this, if

Here we are given the value of

\(\Rightarrow\)

Taking

\(\Rightarrow\)

By calculating