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

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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)