log(2a×3b×5c) = 1/3[log (22×33×5) + log (26×3×57) + log (2×32×54)]
log(2a3b5c) = 1/3 [log(22×32×5×26×3×57×2×32×54]
log(2a3b5c) = 1/3[log(29×36×512)]
log(2a3b5c) = log(23×32×54)
Explanation:-
While it is not explicitly stated that a, b and c are integers, going by the spirit of the question, we are forced to assume that they are integers. In that case, we can equate the powers of 2, 3 and 5 on the LHS and the RHS and say that a= 3, b = 2, and c = 4.