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)#