# Given ln3/ln12 = x and ln5/ln12 = y calculate[log15(400)]/2 .

giorgiana1976 | College Teacher | (Level 3) Valedictorian

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We'll re-write the numbers x and y:

ln 3/ln 12  = log 12 (3) = x and ln 5/ln 12 = log 12 (5) = y

We also re-write the number that has to be calculated:

[log 15 (400)]/2 = log 15 (sqrt 400) = log 15 (20)

We'll add log 12 (3)=x and log 12 (5)=y, we'll get:

log 12 (3) + log 12 (5) = x + y (1)

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

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):

x + y = log 12 (15)

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

1/log 15 (12) = x + y

log 15 (12) = 1/(x + y) (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/(x+y)

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

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

log 15 (5) = y*1/(x+y)

Equating (*) = (**), we'll have:

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+y-x)/(x+y)