Is $x$ rational or irrational when $2^{x}+3^{x}=6$. How to show that?
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I think it should be easy to prove that $2^x$ and $3^x$ are both transcendental, hence $x$ is transcendental, hence $x$ is irrational. – barak manos Feb 03 '16 at 09:44
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6@barakmanos, could you consider leaving an answer proving the transcendance of $2^x$ and $3^x$? I'd be quite interested. – Feb 03 '16 at 10:23
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4related from the same author – Déjà vu Feb 03 '16 at 13:17
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1@barakmanos Transcendence of $2^x$ and $3^x$ doesn't imply transcendence of $x$. It does imply its irrationality though. – Wojowu Feb 03 '16 at 13:52
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It is clear that $x$ can not be integer and not irreducible rational because in this case $2^{\frac nd}$ and $3^{\frac nd}$ are linearly independant over $\mathbb Q$. – Piquito Feb 03 '16 at 17:07
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Schanuel's Conjecture implies $x$ is transcendental. – Mason Sep 15 '22 at 05:24
2 Answers
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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.
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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.
Barry Cipra
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