another log question...

first of all,
log₁₂(60) = log₁₂(15) + log₁₂(4) = log 15 + 2 log 2log 6 + log 2 = k (say)

now, a = {log 15 + log 2}log 6 and, b = {log 6 + 2 log 2}log 15

=> (a + 1) = {log 15 + log 2 + log 6}log 6 and, (b + 1) = {log 6 + 2 log 2 + log 15}log 15

=> k = {(b + 1)log 15 - log 6}{(a + 1)log 6 - log 15}

=> k = {(b + 1)log₆15 - 1}{(a + 1) - log₆15} - - - - (*)

Now comes the calculation for log₆15

clearly we have,

a log 6 - log 15 = log 2 -- (i) and,

b log 15 - log 6 = 2 log 2 --- (ii)

=> b log 15 - log 6 = 2a log 6 - 2 log 15

=> b log₆15 - 1 = 2a - 2 log₆15

=> log₆15 (b + 2) = (2a + 1)

=> log₆15 = (2a + 1)(b + 2)

substituting this value in our star eqn. we get,

k = {(b + 1)(2a + 1) - (b + 2)}{(a + 1)(b + 2) - (2a + 1)}

maybe correct i guess...!!