Let x be a positive transcendental number and n be a positive rational number. Then let y = log(n)/log(x). Clearly, x^{y}=n is a rational. Further, if y were equal to a ratio of integers, a/b, then x^{a}=n^{b}, meaning that x would be an algebraic, contrary to definition. Thus, y is irrational and x^{y} is an example of a power of two irrationals that equals a rational. We could make an even stronger statement like:

Suppose x is algebraic and has some integer power equal to a rational number. Choose the smallest positive integer k such that x^{k}=p is rational. Then, let q be the smallest rational above 1 that p is a power of (or largest rational below one - they'll be reciprocal, so it doesn't matter). Then, as long as n is not an integer power of p, the logarithm log(n)/log(x) will not be rational. For instance, if x = 1/9 * cube root of 4, then k = 3 and x^{k}=p=4/729=2^2*3^-6. Then, q=2*3^-3=2/27, by eliminating common factors in the exponents in the prime factorization (though 2^-1*3^3=27/2 would be equally admissible). So, anything which is not a power of 2/27 would have an irrational logarithm base 1/9 cube root of 4.

This provides us with the machinery to find a lot of examples. If we just take the e^{n} is irrational for any integer n (though this is not a trivial fact; but it could likely be easier to show that e^{n} is never a power of two or some statement like that, which would suffice), it follows that. Thus, if e^{y} is rational, y is irrational. This provides us with lots of examples of irrational^irrational=rational, of the form e^{log(2)}=2.