Can you prove that log ab . log bc . log ca = 1?

Let (log a)/(b-c) = (log b)/(c-a) = (log c)/(a-b) = k, where k is a constant, then log a = kb-kc . (1) log b = kc-ka . (2) log c = ka-kb . (3) Adding (1), (2) and (3), we get log a + log b + log c = 0 log abc = 0 abc = 1 Solution 2: Let (log a)/(b-c) = (log b)/(c-a) = (log c)/(a-b) = k,