Question 274779
To prove something false, we usually resort to a counterexample. A counterexample is an explicit example in which proves a theorem or equation false (usually by a contradiction of some sort).


So let's say that {{{x=b}}} and {{{y=b}}}. This would then mean that {{{log(b,(x))/log(b,(y))=log(b,(b))/log(b,(b))=1/1=1}}}



Now plug {{{x=b}}} and {{{y=b}}} into {{{log(b,(x))-log(b,(y))}}} to get {{{log(b,(x))-log(b,(y))=log(b,(b))-log(b,(b))=1-1=0}}}



So in short, {{{log(b,(x))/log(b,(y))=1}}} and {{{log(b,(x))-log(b,(y))=0}}} when {{{x=b}}} and {{{y=b}}}.



Clearly {{{1<>0}}} which means that {{{log(b,(x))/log(b,(y))<>log(b,(x))-log(b,(y))}}} when {{{x=b}}} and {{{y=b}}}.


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Here's another way of looking at it.



By the change of base formula, {{{log(b,(x))/log(b,(y))=log(y,(x))}}}


By another identity, {{{log(b,(x))-log(b,(y))=log(b,(x/y))}}}



So let's assume that {{{log(b,(x))/log(b,(y))=log(b,(x))-log(b,(y))}}}. If this is the case, then {{{log(y,(x))=log(b,(x/y))}}}



Equate the bases and arguments to get {{{y=b}}} and {{{x=x/y}}}. The second equation simplifies to {{{1=1/b}}}. Solve for b to get {{{b=1}}}. 



Now if {{{b=1}}}, this means that {{{log(b,(x))=log(10,(x))/log(10,(b))=log(10,(x))/log(10,(1))=log(10,(x))/0}}} which is impossible (you can't divide by zero).