TITLE: How do you prove that integer powers of $1.5$ are not equal to any integer powers of $2$?
QUESTION [7 upvotes]: I've read somewhere that no integer power of $1.5$ can ever equal any integer power of $2$ (besides the zeroth power, of course). It makes sense, but how is this proven?
(The application here is music. Octaves have a frequency ratio of $2:1$. Pure $5$ths have a $3:2$ frequency ratio. This issue—octaves never lining up with fifths—leads to either the compromise of temperaments or awful sounding intervals that prevent modulation.)

REPLY [2 votes]: More generally: if an integer power $\,n>0\,$  of a fraction $\,q\,$ is an integer $\,a,\,$ then  $\,q\,$ is an integer, by the Rational Root Test, i.e. $\,q\,$ is a root of the $\rm\color{#c00}{monic}$ $\,f(x) \:\!=\:\! \color{#c00}{1}\:\!x^n - a\,$ so writing $\,q = c/d\,$ in least terms, RRT $\Rightarrow d\mid \color{#c00}1,\,$ so  $\,d=1,\,$ so $\,q=c/d\in\Bbb Z$.
Generally RRT shows that a fractional root of any integer coefficient polynomial $\,f(x)\neq 0\,$ with $\color{#c00}{\rm lead\ coeff= 1}\,$ must  be an integer. Above is the special case of $f$ being binomial (two terms).
Remark $ $ The arguments in the other answers are essentially special cases of the (short and simple) proof of RRT. The above is but a glimpse of the key role that RRT plays in  factorization and number theory - which is clarified when one studies more general number rings (e.g. domains with unique prime factorization must be integrally closed, i.e. they must satisfy said $\color{#c00}{\rm monic}$ case of RRT).