SOLUTION: Prove (disprove) that there exists an A and B such that logA(B) = logB(A) where A ǂ B, both A and B > 0, and both A and B ǂ 1; note that a specific example is not a pro

Question 1010275: Prove (disprove) that there exists an A and B such that logA(B) = logB(A) where A ǂ B, both A and B > 0, and both A and B ǂ 1; note that a specific example is not a proof

Answer by Edwin McCravy(20056) (Show Source): You can put this solution on YOUR website!

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

which is true if and only if

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

By the definition of logarithms, that is equivalent to

We cannot use the + for that would make A = B, which is
not allowed.
So the above is true if and only if

and A ǂ B, both A and B > 0, and both A and B ǂ 1
Edwin
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SOLUTION: Prove (disprove) that there exists an A and B such that logA(B) = logB(A) where A &#450; B, both A and B > 0, and both A and B &#450; 1; note that a specific example is not a pro