Question 1010275
{{{log(A,(B))}}}{{{""=""}}}{{{log(B,(A))}}}

Use the change of base formula, we have that the above 
is true if and only if: 

{{{log(A,(B))}}}{{{""=""}}}{{{1/log(A,(B))}}}

which is true if and only if

{{{(log(A,(B)))^2}}}{{{""=""}}}{{{1)}}}

Use the principle of square roots, we have that the above
is true if and only if

{{{log(A,(B))}}}{{{""=""}}}{{{"" +- sqrt(1)}}}

By the definition of logarithms, that is equivalent to

{{{A}}}{{{""=""}}}{{{B^("" +- 1)}}}

We cannot use the + for that would make A = B, which is
not allowed.

So the above is true if and only if 

{{{A}}}{{{""="""}}}{{{B^(-1)}}}

{{{A}}}{{{""=""}}}{{{1/B}}}

{{{AB}}}{{{""=""}}}{{{1}}}

and A ǂ B, both A and B > 0, and both A and B ǂ 1

Edwin</pre>