log a (x*y) = log a (x) + log a (y)
The derivation of this law is a bit trickier than the first two. Firstly, we need to relate x and y to the
base a. So, assume that x = a ^ (m) and y = a ^ (n).
log a (x) = m
log a (y) = n
This means that we can write:
log a (x*y) = log a ( a^(m) . a^(n) )
= log a ( a ^ (m+n) )
= log a ( a ^ ( log a (x)+log a (y)
)
= log a (x) + log a (y)
For example, show that
log ( 10 * 0.1 ) = log 10 + log 0.1
= log ( 1 )
= log ( 10 ^ (0) )
= 0
log ( 10 * 1 ) = log 10 + log 1
= log ( 10 )
= log ( 10 ^ (1) )
= 1
log ( 10 * 10 ) = log 10 + log 10
= log ( 100 )
= log ( 10 ^ (2) )
= 2
log ( 10 * 100 ) = log 10 + log 100
= log ( 1000 )
= log ( 10 ^ (3) )
= 3
