Mathematics-Logarithms

Logarithm in mathematics-The word Logarithm is made up of 2 Greek words.
        ( 1 ) logos ( means " ratio " ) and 
        ( 2 ) arithmos ( means " number " ).

The concept of ' logarithms ' was introduced by Scottish Mathematician John Napier ( 1550 - 1617 ).

Henry Briggs ( 1556 - 1630 ) introduced the decimal logarithms.

Let us explain via an example;

⇒ 3³ = 27
⇒ Apply log on both sides.
⇒ log ( 3³ ) = log 27.
⇒ 3 log 3 = log 27.
⇒ 3 = ( log 27 ) / ( log 3 ).
⇒ 3 = log ₃27.
∴ in general, for any positive real number a, a ≠ 1
     a^x = m ⇔ log _am = x

Definition :: -

Let a be a Positive real number a ( a ≠ 1 ), and x be the unique real number such that a^x = m, for a positive real number m, then we say that logarithm of m to the base a is x or x is logarithm of m to the base a, written as

log _am = x.

Consider the properties :

1 ) For any positive real number a, a ≠ 1, a⁰ = 1 ⇔ log _a1 = 0
2 ) For any positive real number a, a ≠ 1, a¹ = a ⇔ log _aa = 1
3 ) For any positive real number a, a ≠ 1, a^x = a^x ⇔ log _aa^x  = x

Laws of Logarithms ::-

1 ) Product Rule ( First Law )

⇒The logarithm of the product of two numbers is equal to the sum of the logarithms of the numbers, with reference to the same base.
i.e., log _a( mn ) = log _am + log _an

⇒Proof :: -
log _am = x and log _an = y
Then m = a^x and n = a^y
mn = a^x × a^y ⇔ a^{x + y},

Hence log _a( mn ) = x + y = log _am + log _an

2 ) Quotient Rule ( Second Law ) 

⇒ The logarithm of a quotient is equal to the logarithm of the numerator minus the logarithm of the denominator.
i.e., log _a( m/n ) = log _am - log _an .

⇒ Proof ::-
Let  log _am = x and log _an = y,
Then m = a^x and n = a^y
m / n = a^x / a^y ⇔ a^x-y
Hence log _a( m / n ) = x - y ⇔ log _am - log _an.

3 ) Power Rule ( Third Law ) 

⇒The logarithm of the n^th power of a number is n times the logarithm of the number.
i.e., log _a( mⁿ ) = n × log _am 

⇒ Proof ::-
Let log _am = x, so that m = a^x
∴mⁿ = ( a^x )ⁿ = a^xn

Hence log _a( mⁿ ) = nx = n × log _am.

4 ) Base changing Rule ::-

⇒ log _bm = log _am / log _ab 

⇒ Proof ::-
Let x = log _bm,
We have b^x = m,
Taking logarithm to the base a on both sides, we get log _a( b^x ) = log _am 
i.e., x log _ab = log _am / log _ab
i.e, log _bm = log _am / log _ab