Is $x$ irrational when $2^{x}+3^{x}=6$?

Is $x$ rational or irrational when $2^{x}+3^{x}=6$. How to show that?

Solution 1:

Suppose $x=\frac{m}{n}$ is rational for $n,m\in\Bbb N$. It's clear that $2^x=\sqrt[n]{a}$ and $3^x\sqrt[n]{b}$ are non-integers, where $a=2^m,b=3^m$, hence are irrational (it's an elementary result that $n$-th root of an integer is either an integer or is irrational). By positive answer to this question it follows that a sum of irrational roots of positive integers is irrational, hence can't be 6. This is a contradiction.

Hence $x$ is irrational.

Actually, we don't need all of the strength of the theorem from the post above. Since $2,3$ are relatively prime, the result from this paper, which is completely elementary, is applicable.

Solution 2:

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.