Proving a^x = a^y iff x = y

Date: 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/