How that step (marked in pencil) is derived ? Question is in the comment box

51. logb a log. a - log, a og a log a log b loggc Let log a= A, log b = B and log c = C log a (log b) (log c) -1- Then, log, a loge a-log, a = A2 -1 BC B2 Similarly, loga b log. b- logp b= 1 CA and loga C log, C -loge C = - -1 AB (log, a log, a- loga a) + (logą b loge b - log, b) +(l09, C log, C - log. C) = 0 2 A-1+5-1+ C 1 0 A2 BC CA AB or A3 +B +C3 3ABC = 0 A+B+C = 0 log a +log b + log c =0 log (abc) =0 abc