Proving a^x = a^y iff x = yDate: 12/13/2000 at 12:35:58 From: Lauren Subject: Proving logarithmic properties I am trying to prove that a^x = a^y iff y = x. I have proved (=>) by contradiction, for the case when x and y are natural numbers. How do I prove this for all x and y? Date: 12/13/2000 at 15:34:24 From: Doctor Rob Subject: Re: Proving logarithmic properties Thanks for writing to Ask Dr. Math, Lauren. You must be assuming that a is a positive real number, but not equal to 1. First prove (=>) for negative numbers: Assume a^(-x) = a^(-y) for x and y natural numbers. Then multiply both sides of the equation by a^(x+y): a^y = a^x This implies by what you already proved that y = x, so -x = -y. Next prove (=>) for rational numbers. Suppose: a^(r/s) = a^(t/u) where r, s, t and u are integers. Now raise both sides to the power s*u: a^(r*u) = a^(t*s) Now use the result for integers to conclude that: r*u = s*t r/s = t/u so the rational exponents must be equal. Now to prove it for all real numbers, you can use the fact that every real number is the limit of a sequence of rational numbers, and the fact that the function x |--> a^x is continuous. Alternatively, you can use the Dedekind cut construction of the real numbers from the rationals to show that the cuts for a^x and a^y being equal implies that the cuts for x and y are equal, using the result above for rational numbers. - Doctor Rob, The Math Forum http://mathforum.org/dr.math/ |
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