#LOGARITHM-1

If log (20) = A ; log (45) = B ; log (24) = C, what is the value of  log (120)?
(a) (3A + 2C + B)/4
(b) (3B + 2A + C)/4
(c) (3C + 2B + A)/4
(d) (3B - 2A + C)/4

Solution-

log20 = log 4*5 = log 4 + log 5 = 2log 2 + log 5 = A ---------(i)
Similarly log 45 = log 9 + log 5 = 2log 3 + log 5 = B--------(ii)
log 24 = log 8 + log 3 = 3log 2 + log 3 = C ---------(iii)
Now, log 120 = log (3*5*8) = 3log 2 + log 3 + log 5

Now, all the options are divided by 4,so let's multiply (log 120) by 4
we get, 4* log 120 = 12*log 2 + 4*log 3 + 4*log 5-------(iv)
log 3 is in eqn (ii) and (iii), so if we multiply eqn (iii) by 2 and add eqn (ii),we get,
2log 3 + log 5  + 6log 2 + 2log 3 = B + 2C-------(v)
=>6log 2+ 4log 3 + log 5 = B + 2C
Now if we add 6log 2 + 3log 5, we will get the required result
Multiply eqn (i) by 3,
6log 2 + 3log 5 = 3A----------(vi)
 Adding eqn(v) and (vi), we get4* log 120 = 12*log 2 + 4*log 3 + 4*log 5 = 3A + 2C + B
Hence, option (a) is the answer.

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