Prove that Log b (a) + Log c (b) + Log a (c) = 1/Log a (b) + 1/Log b (c) + 1/Log c (a)

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So, the rule I am going to use here is that Log x (y) = (Log y)/(Log x)Log b (a) + Log c (b) + Log a (c) = (Log a)/(Log b) + (Log b)/(Log c) + (Log c)/(Log a)Now we use the inverse of that rule, that is (Log x)/(Log y) = 1/ (Log y/ Log x) = 1/Log x (y)(Log a)/(Log b) + (Log b)/(Log c) + (Log c)/(Log a) = 1/(Log a (b)) + 1/(Log b (c)) + 1/(Log c (a))Hope that makes sense!!