Rule 1
The logarithm of 1 to any base is 0.
Proof
We know that any number raised to zero equals 1. That is, a^{0} = 1, where "a" takes any value. Therefore, the logarithm of 1 to the base a is zero. Mathematically, we express this as log_{a}1 = 0.
Example
What is the value of log_{10}1.
Needless to say this would be zero.
Rule 2
The logarithm of a number where the number is the same as the base is 1.
Proof
We know that any number raised to the power of 1 is itself. That is a^{1} = a. Therefore, the logarithm of a to the base a is equal to 1.
Mathematically, we express this as log_{a}a = 1.
Example
What is the value of log_{13}13?
By applying the above rule, the value of log_{13}13 is 1.
Rule 3
The logarithm of a product to base a is equal to sum of the logarithms of the individual numbers which constitute the product to the same base a. That is, log_{a}M.N = log_{a}M + log_{a}N.
Proof
If M.N is the product and if a^{x} = M and a^{y} = N, then M.N = a^{x} . a^{y}.
By the law of indices a^{x}. a^{y} = a^{x+y}. Therefore,
a^{x+y} = M.N
Then the logarithm of M.N to base a is equal to x + y. Mathematically, it will be
log_{a} M.N = x + y ......(1)
Now, if we express a^{x} = M and a^{y} = N, in terms of logarithms they will be log_{a} M = x and log_{a} N = y. Substituting the values of x and y in 1, we have
log_{a} (M.N) = log_{a} M + log_{a} N
Example
What is the value of log_{3}33?
We know that 33 can be expressed as the product of 3 and 11. That is, log_{3} 33 = log_{3} (3 x 11). Applying the above rule this can be expressed as log_{3} 3 + log_{3} 11. Since log_{3}3 is 1, we rewrite it as log_{3} 33 = 1 + log_{3} 11.
Rule 4
The logarithm of a fraction to the base a will be equal to the difference of the logarithm of the numerator to the base a and the logarithm of the denominator to base a. That is, log_{a} (M/N) = log_{a} M - log_{a} N.
Proof
Let a^{x} = M and a^{y} = N. Then M/N = a^{x}/a^{y}. By the law of indices, this will equal to a^{x-y}. The logarithm of M/N to base 'a' will, therefore, be x - y. Mathematically this is expressed as
log_{a} (M/N) = x - y .......(1)
If we express a^{x} = M and a^{y} = N in terms of logarithms, they will be log_{a} M = x and log_{a} N = y. Substituting the values of x and y in (1), we have
log_{a} (M/N) = log_{a} M - log_{a} N
Example
What is the value of log_{2} (32/4).
By applying the above rule, this can be written as log_{2} 32 - log_{2} 4. This can be further solved. But we look at it only after learning the next rule.
Rule 5
The logarithm of a number raised to any power, integral or fractional, is equal to product of that number and the logarithm of the number raised to base a. That is, log_{a} (M^{P}) = p.log_{a}M.
Proof
If M = a^{x}, then log_{a} M = x. Now suppose that M is raised to the power of n, that is M^{n}. Since M = a^{x}, M^{n} = a^{nx}. This is in accordance with the priniciple that if we perform any operation on an equation it should be performed on both the sides of the equation in order to keep the equation symbol valid.
M^{n} = a^{nx}, if expressed in terms of logarithms will be
log_{a}(M^{n}) = nx ...........(1)
On substituting log_{a} M = x in (1), we have
log_{a} (M^{n}) = n . log_{a} M
Similarly if n = 1/r, we have
log_{a} (M^{1/r}) = (1/r) . log_{a} M
Now we take up the example discussed under Rule 4 and look at how it is further simplified. Before we go on to the next step, let us express log_{2} 32 and log_{2} 4 as log_{2} 2^{5} and log_{2} 2^{2}. By rule 5, these are expressed as 5.log_{2}2 and 2.log_{2} 2. And since log_{2} 2 is one, 5.log_{2}2 and 2.log_{2}2 reduce to 5.1 = 5 and 2.1 = 2. Therefore, log_{2} 32 - log_{2}4 when simplified gives
log_{2}(2^{5}) - log_{2}(2^{2})
= 5.log_{2}2 - 2.log_{2}2
= 5.1 - 2.1
= 5 - 2 = 3.
We obtain the same value even by simplifying the term on the left hand side. We know that 32/4 = 8. That is, log_{2}8 can be expressed as 2^{3}. On application of rule 5, this will be 3.log_{2} 2. Again this gives us 3.1 = 3.
Generally, logarithms are expressed to base 10 and base 'e'. While the logarithms expressed to base 10 are referred to as common logarithms, those expressed to base 'e' are referred to as Napier or Natural logarithms. The value of 'e' is approximately 2.718. In practise common logarithms are expressed as 'log' while natural logarithms are expressed as 'ln'. We want to emphasize that generally the base is not stated and by looking at the manner it is expressed we ought to decide whether it is a common or natural logarithm.