Suppose that we know the logarithms of all numbers which are expressed to base 'a' and we are required to find the logarithms of all these numbers to base 'b'. We proceed as follows. Let N be any one of the numbers of which we are required to find the logarithm to base 'b' and the value itself be some 'x'. That is, log_{b}N = x or N = b^{x}. But we already know the value of log_{a} N. Also log_{a}N can be expressed as log_{a}(b^{x}) as N = b^{x}. By rule 5, log_{a}(b^{x}) can be expressed as x.log_{a}b or
xlog_{a}b = log_{a}N
...............(1)
Since the values of N and b are known, the values of log_{a} N and log_{a} b can be found from the tables. These values when substituted in equation (1) gives us the value of log_{b} N.
In the above equation, what will happen if N = a. Equation (1) will be

(because log_{a}a = 1)

or
log_{b}a x log_{a} b = 1
Example
Find the values of the following.

log_{3} 81
We know that 81 = 3^{4}. Therefore, log_{3} 3^{4} = 4. log_{3} = 4.1 = 4

log_{3} (9 x 27 x 81)
log_{3} (9 x 27 x 81) = log_{3}9 + log_{3}27 + log_{3}81
(log_{a} M.N = log_{a} M + log_{a}N)
= log_{3}(3^{2}) + log_{3}(3^{3}) + log_{3}(3^{4})
= 2.log_{3}3 + 3.log_{3}3 + 4.log_{3}3
= 2 + 3 + 4 = 9

In this example we apply an extension of rule 5.
(5/4).log_{3}3 = (5/4).1 = (5/4)

log_{3} (243/81)
log_{3}243  log_{3}81 [log_{a} (M/N) = log_{a}M  log_{a} N]
log_{3}3^{5}  log_{3}3^{4}
5.log_{3}3  4.log_{3}3 = 5  4 = 1
(log_{M} (M^{P}) = p.log_{M} M = p)