I need this to be true to prove something else.

3^(1.6) = 2^(2.53594+)

3 is odd, 2 is even, both exponents are real numbers as requested.

The theorem is false.

If there are more constraints, please let us know.
Sorry, can m and n be any real numbers? If so, then it's just not true.

Quick notational admin to avoid using subscripts: I'm going to denote

>the logarithm of x, with respect to base y

as

>log$x,y$.

For example, I would write ln(3) as log$3,e$ and 3 could be considered equal to log$1000,10$. Now, with that out of the way...

Take a fixed positive real number R. For fixed a and b, let m = log$R,a$ and n = log$R,b$. Then, we see:

* a^(log$R,a$) := R,
* b^(log$R,b$) := R.

These values exist (and are real) for all real positive R, so we can let R be anything, not just 1. Which in turn means that m and n could be anything, not just 0.

If you wanted m and n to be integers (or even just rational) then you have some stuff to work with thanks to the Fundamental Theorem of Arithmetic. But if they can be any real numbers, you're out of luck.

**Edit:** Final paragraphs mixed a and b up with m and n. Have fixed it. Conclusion remains the same.
m&n's are real numbers? Little equivalence classes of Cauchy sequences of rational numbers in a hard candy shell. And they're delicious.

By the way, OP, we need specifically that m and n are integers.

In this case, if a is even and m is nonzero, then that means that 2 divides a, which means that 2 divides a\^m. This means that 2 also divides b\^n, and since 2 is prime, it divides b.

If m and n are reals, then you can easily find counter examples using logarithms.