Definition: If a is a positive number, other than 1 and x is a rational number such that a^x = n , then we say that logarithm of n to the base a is x . It is written as log_an = x .

i.e. a^x = n \Longleftrightarrow log_an = x

Some examples for you to grasp this concept a little more:

2^5 = 32 \Longleftrightarrow \log_2{32} = 5

3^4 = 81 \Longleftrightarrow \log_3{81} = 4

10^3 = 1000 \Longleftrightarrow \log_{10}1000 = 3

81^{\frac{1}{4}} = 3 \Longleftrightarrow \log_{81}3 = \frac{1}{4}

7^0 = 1 \Longleftrightarrow \log_71 = 0

10^{-2} = 0.01 \Longleftrightarrow \log_10{0.01} = -2

Note: For any positive real number a, we know that a^0 = 1 . Hence \log_a1 = 0 and \log_aa = 1 and

Fundamental Laws of Logarithms:

First Law:

If m, \ n are positive rational numbers, then \log_a(mn) = \log_a m + \log_a n i.e. the logarithms of the product of two numbers is equal to the sum of their logarithms.

We can use this law and extrapolate it to multiple situations like:

  • If m, n, p are positive numbers, then \log_a(mnp) = \log_a m + \log_a n + \log_a p
  • If x_1, x_2, x_3, ... , x_n are positive numbers, then \log_a(x_1.x_2.x_3...x_n) = \log_a x_1 + \log_a x_2 + \log_a x_3 \cdots + \log_a x_n
  • If x, y are real numbers such that xy>0 , then \log_a(xy) = \log_a |x| + \log_a |y| 

Example 1: Prove \log_{10} (1+2+3) = \log_{10} 1 +  \log_{10} 2 +  \log_{10} 3 

LHS = \log_{10} (1+2+3) = \log_{10} (6) = \log_{10} (1\times 2 \times 3) 

 = \log_{10} 1 +  \log_{10} 2 +  \log_{10} 3  = RHS

Example 2: Prove \log_{10} 2 + 1 = \log_{10} 20

LHS = \log_{10} 2 + 1

 = \log_{10} 2 + \log_{10} 10 = \log_{10} (2 \times 10) = \log_{10} 20 = RHS

Second Law:

If m, \ n are positive rational numbers, then  \log_a ( \frac{m}{n} ) = \log_a m - \log_a n  i.e. the logarithms of the ratio of two numbers is equal to the difference of their logarithms.

The above law can also be stated as follows: If x, y are real numbers such that xy>0 , then \log_a ( \frac{x}{y} ) = \log_a |x| - \log_a |y|

Example 1: Evaluate:

\log_{10} 500 - \log_{10} 5 = \log_{10} ( \frac{500}{5} ) = \log_{10} 100 = 2 \log_{10} 10 = 2

Example 2: Evaluate:

\log_6 72 - log_6 2 = \log_6 ( \frac{72}{2} ) = \log_6 36 = 6 \log_6 6 = 6

Third Law:

If m, \ n are positive rational numbers, then \log_a(m^n) = n. \log_a m

Example 1: \log_2(3^2) = 2. \log_2 3

Example 2: \log_3(2^2) = 2. \log_3 2

Fourth law:

\log_a 1 = 0 i.e. the log of 1 to any base is always 0 .

Fifth law:

\log_a a = 1 i.e. the log of any positive number to the same base is always 1 .

Example 1: Evaluate:

\frac{\log_a 125}{\log_a \sqrt{5}} = \frac{\log_a 5^3}{\log_a 5^{1/2}} = \frac{3 \log_a 5}{\frac{1}{2} \log_a 5} = 6

Example 2: Evaluate:

\log_2 ( \log_2 ( \log_2 16)) =  \log_2 ( \log_2 ( \log_2 2^4))

  =  \log_2 ( \log_2 4) = \log_2 ( \log_2 2^2) = \log_2 2 = 1

Sixth law (Base change formula):

If m is a positive rational number and a, b are positive real numbers such that a \neq 1 and b \neq 1 , then \log_a m = \frac{\log_b m}{\log_b a} .

In the above formula if b = m, then we get \log_a m = \frac{\log_m m}{\log_m a} =  \log_a m . \log_m a = 1

Example 1: Evaluate:

\log_b a . \log_c b . \log_a c  = (\log_b a. \log_c b). \log_a c = \log_c a . \log_a c = 1

Since \log_b a = \frac{\log_c a}{\log_c b} we get

Example 2: Evaluate:

\frac{1}{\log_2 42} +  \frac{1}{\log_3 42} +  \frac{1}{\log_7 42} 

= \log_{42} 2 + \log_{42} 3 + \log_{42} 7

= \log_{42} (2 \times 3 \times 7) = \log_{42} 42 = 1

Seventh Law:

If a is a positive real number and n is a positive rational number, then a^{\log_a n} = n

Example 1: 3^{\log_3 8} = 8

Example 2: 2^{3\log_2 5} = 2^{\log_2 5^3} = 5^3

Eighth Law:

If a is a positive real number and  n is a positive rational number, then \log_{a^n} n^p = \frac{p}{q} \log_a n

Example 1: \log_{81} 243 = \log_{3^4} 3^5 = \frac{5}{4} \log_3 3 = \frac{5}{4}

Example 2: \log_{1024} 64 = \log_{2^{10}} 2^6 = \frac{6}{10} \log_2 2 = \frac{3}{5}

Ninth Law:

If x, y and a(\neq 1) are positive real numbers, then x^{\log_a y} = y^{\log_a x}

Example 1: 16^{\log 3} = (4^2)^{\log 3} = 4^{\log 3^2}= 4^{\log 9} = 9^{\log 4}

Example 2: 27^{\log 2} = (3^3)^{\log 2} = 3^{\log 2^3} = 3^{\log 8} = 8^{\log 3}

As you would have noticed, we have listed all the key concepts of logarithms (as related to Class 9 students). We have also presented a couple of example / applications of the laws. However, we would recommend the student to solved all the problems as stated in Exercise 8(a). More you practice, the better.

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