Prime factorization and real exponents

[e2m2a]

TL;DR

Does the unique factorization theorem apply when prime numbers are raised to any real number power?

I know that the prime factorization theorem predicts that a prime number raised to an integer power will never be equal to another prime number raised to a different power. But does this apply to real number powers? For example, suppose there is a prime number raised to some real value, could it be equal to another prime number raised to a different real value?

 

[fresh_42]

Not sure what you mean. But the answer is: there are no prime numbers in the reals. Prime elements are certain elements of a ring. They cannot be units. But every real number different from zero is a unit, hence no real primes. Of course you can solve any equation $a^x=b^y$, but this has nothing to do with primes.

 

[PeroK]

Summary:: Does the unique factorization theorem apply when prime numbers are raised to any real number power?

I know that the prime factorization theorem predicts that a prime number raised to an integer power will never be equal to another prime number raised to a different power. But does this apply to real number powers? For example, suppose there is a prime number raised to some real value, could it be equal to another prime number raised to a different real value?

Yes, of course. In general, the powers of a prime form a continuous function $p^x$, that takes every value from $p^0 = 1$ upwards.

Take any real number, $y > 1$, and any prime $p$, then:
$$p^{\frac{\ln y}{\ln p}} = y$$
In other words, $y$ can be expressed as a power of any prime.

 

[e2m2a]

ok thanks for the reply