Is it possible to have $x=2^N$ if $N$ is not integer and both $x$ and $N$ are rational numbers?

Question

Is it possible to have $x=2^N$ if $N$ is not integer and both $x$ and $N$ are rational numbers?

If possible, give an example of $x$ and $N$. If not possible, explain why.

Answer 1

No, it is not possible. Assume that $a/b=2^{c/d}$ where $\gcd(a,b)=1$, $\gcd(c,d)=1$, and $d>1$. Then $a^d=2^c b^d$. Hence $a^d$ is even which implies that also $a$ is even and therefore $b$ is odd because $(a,b)=1$. Let $a=2^s k$ where $s\geq 1$ and $k$ is odd, then $2^{sd} k^d=2^c b^d$. Since the integer factorization is unique, we have that $sd=c$ and $\gcd(c,d)=\gcd(sd,d)=d>1$. Contradiction.

Answer 2

Hint:

No it is not. put $x=\frac{c}{d}$ and $N=\frac{a}{b}$ then we should have $2^a=\frac{c^b}{d^b}$ you can play with this equality and conclude that it is not possible.

Answer 3

Suppose $N=p/q$ where $p,q\in\mathbb{Z}$. Then $$ x=2^{p/q} $$ and thus, $$ x^q-2^p=0 $$ By the Rational Root Theorem, $x\in\{\pm1,\pm2\}$. None work unless $p=q$ or $p=0$, but this would mean $N$ is an integer.