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

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> 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      Log On


   

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) About Me  (Show Source):
You can put this solution on YOUR website!
log%28A%2C%28B%29%29%22%22=%22%22log%28B%2C%28A%29%29
Use the change of base formula, we have that the above
is true if and only if:
log%28A%2C%28B%29%29%22%22=%22%221%2Flog%28A%2C%28B%29%29
which is true if and only if
%28log%28A%2C%28B%29%29%29%5E2%22%22=%22%221%29
Use the principle of square roots, we have that the above
is true if and only if
log%28A%2C%28B%29%29%22%22=%22%22%22%22+%2B-+sqrt%281%29
By the definition of logarithms, that is equivalent to
A%22%22=%22%22B%5E%28%22%22+%2B-+1%29
We cannot use the + for that would make A = B, which is
not allowed.
So the above is true if and only if
A%22%22=%22%22%22B%5E%28-1%29
A%22%22=%22%221%2FB
AB%22%22=%22%221
and A ǂ B, both A and B > 0, and both A and B ǂ 1
Edwin