The Change of Base Formula merely states that:
#log_color(green)(b) color(red)(a) = (log color(red)(a))/(log color(green)(b)) = (ln color(red)(a))/(ln color(green)(b))#
(By the way, the equivalence to #ln a/ln b# just means that #lnx/logx# is a constant: #~~2.303#.)
You just end up taking the base #10# or #e# logarithm of #a# and divide it by the same kind of #log# on #b#. The base changes so that both bases are the same. For example:
Base 10 counting (decimal system)
#1, 2, 3, 4, 5, 6, 7, 8, 9, 10, ...#
Base 2 counting (binary system)
#1, 10, 11, 100, 101, 110, 111, 1000, 1001, 1010, ...#
#1010# in base 2 (you can emphasize this as #1010_2#) is equal to #10# in base 10 (you can emphasize this as #10_10#).
To get two people to communicate well, they must speak the same language. To be able to divide two logarithms at all, you have to get the numbering systems to be the same.
Conveniently, since #(log a)/(log b) = (ln a)/(ln b)#, you don't even need to specify the base of the "new #log#". All you do is:
#log_2 15 = (log 15)/(log 2) = ln 15 / ln 2 = (ln (3*5))/ln 2 = color(blue)((ln 3 + ln 5) / ln 2 ~~ 3.907)#
