How do you prove that integer powers of $1.5$ are not equal to any integer powers of $2$?

Question

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.)

Answer 1

Apart from $1.5^0$, no integer power of $1.5$ is an integer. We have $1.5=\frac32$, and the fundamental theorem of arithmetic shows that $\frac{3^n}{2^n}$ can never be simplified.

Answer 2

You know that $1.5=\frac{3}{2}$.

To have a power of 1.5 equal to a power of 2 means that $$\left(\dfrac{3}{2}\right)^m=2^n$$ for some integers $m,n\in\mathbb Z$.

This means $2^{m+n}=3^m$. Since $2$ and $3$ are prime numbers, this can't hold unless $m=n=0$ (see the fundamental theorem of number theory).

Answer 3

If $\left(\frac32\right)^n=2^m$, then $3^n=2^{m+n}$. For all but trivial cases, the right hand side is even, whereas the left is always odd.

Answer 4

We have $$1.5^n=\left(\frac{3}{2}\right)^n=\frac{3^n}{2^n}$$ For this to equal a power of 2 we need $$\frac{3^n}{2^n}=2^k$$ $$3^n=2^k \cdot 2^n = 2^{(n+k)}$$ And as no power of 3 is equal to a power of 2 ($2^a$ and $3^b$ are relatively prime), we cannot have $1.5^a=2^b$.

Answer 5

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).