If log 12 (3)=a and log 12 (5)=b, what is log 15 (20)=?

We notice that if we'll add log 12 (3)=a and log 12 (5)=b, we'll
get:

log 12 (3) + log 12 (5) = a + b
(1)

Since the bases are matching, we'll apply the rule of
product:

log 12 (3) + log 12 (5) = log 12
(3*5)

log 12 (3) + log 12 (5) = log 12 (15)
(2)

We'll substitute (1) in (2):

a + b
= log 12 (15)

But log 12 (15) = 1/log 15
(12)

1/log 15 (12) = a + b

log 15 (12)
= 1/(a+b) (3)

log 15 (12) = log 15
(4*3)

log 15 (4*3) = log 15 (4) + log 15
(3)

log 15 (4) = log 15 (12) - log 15 (3)
(*)

Now, we'll calculate log 15
(20):

log 15 (20) = log 15 (4*5)

log 15
(4*5) = log 15 (4) + log 15 (5)

log 15 (4) = log 15
(20) - log 15 (5) (**)

We'll write log 15 (3) with
respect to log 12 (3):

log 15 (3) = log 12 (3)*log 15
(12)

log 15 (3) = a*1/(a+b)

We'll write
log 15 (5) with respect to log 12 (5):

log 15 (5) = log 12 (5)*log
15 (12)

log 15 (5) = b*1/(a+b)

We'll
put (*) = (**):

log 15 (12) - log 15 (3) = log 15 (20) - log 15
(5)

We'll add log 15 (5) both
sides:

log 15 (20) = log 15 (12) - log 15 (3) + log 15
(5)

log 15 (20) =
(1+b-a)/(a+b)