TITLE: Is $x$ irrational when $2^{x}+3^{x}=6$?
QUESTION [11 upvotes]: Is $x$ rational or irrational when $2^{x}+3^{x}=6$. How to show that?

REPLY [2 votes]: If $\gcd(m,n)=1$ with $n\gt1$ and we let $u=2^{m/n}$, the minimal polynomial for $u$ is of degree $n$, namely $P(u)=u^n-2^m=0$.  But if $2^{m/n}+3^{m/n}=6$, then
$$3^m=(6-u)^n=6^n-{n\choose1}6^{n-1}u+\cdots+(-1)^nu^n$$
which we can rewrite as
$$Q(u)=(-1)^nu^n+\cdots-{n\choose1}6^{n-1}u+6^n-3^m=0$$
Adding or subtracting $P$ and $Q$ to cancel the $u^n$ term gives a polynomial of lower degree with $u$ as a root, contradicting the assumed minimality of $P$.  Thus the equation $2^x+3^x=6$ is not satisfied by any rational value of $x$ (since $2^m+3^m\not=6$ for any integers $m$).
Note, the proof is simple only because we're only considering the sum of two rational powers.  An equation like $2^x+3^x+5^x=30$ would (I think) require the more general results referenced in Wojowu's answer.