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Does $6^x+3^x=10$ have a solution that is an algebraic number? I imagine not, but how would one go about proving such a thing?

Deepak
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Johan
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    Edited your post to make it as clear as it should've been at the outset. – Deepak Oct 08 '14 at 07:50
  • There is a solution and it is irrational, but that is all I can say. Proof: Let $f(x) = 6^x + 3^x - 10$. The function $f$ is continuous as it is a sum of continuous functions, and clearly $f$ is monotonically increasing (bounded below by $-10$). Thus $x\in\mathbb R$ such that $f(x) = 0$ exists and is unique. Denote this constant $Z$. Assume by contradiction that $Z\in\mathbb Q$. If $Z = \frac p q$ with $p\in\mathbb Z$, $q\in\mathbb N$, then $6^p + 3^p = 10^q$, which is impossible because when the LHS is an integer it is necessarily odd while the RHS is necessarily an even integer. – Ben Frankel Oct 08 '14 at 10:42
  • @BenFrankel, I do not understand your argument that it can't be rational. How do you conclude that $6^p+3^p=10^q$? That is not what you get if you raise both sides to the power of $q$. – Johan Oct 09 '14 at 06:10
  • @Johan, you are correct, somehow I fell for the Freshman's Dream xS. It is actually rather difficult to prove that $Z$ is irrational, if it even is (which does seem likely). – Ben Frankel Oct 09 '14 at 10:38

2 Answers2

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Wanted to add this as a comment as it is somewhat tangential to the question, but it's too long for the comment box.

It is actually fairly easy to prove that solutions to equations like $6^x = 10$ are necessarily transcendental. First establish irrationality. Assume to the contrary that $x (\in \mathbb{Q}) = \frac{p}{q}$ where $p>q$ and $(p,q)=1$. Prime factorise both sides: $3^p\cdot 2^p = 2^q \cdot 5^q \implies 3^q \cdot 2^{p-q} = 5^q$ which is a contradiction by the uniqueness of prime factorisation. So $x \not \in \mathbb{Q}$. Now you have to use Gelfond-Schneider. $6$ is algebraic (and not $0$ or $1$), $x$ is irrational. If $x$ were algebraic, then $6^x$ would be transcendental, but clearly, $10$ is not. Therefore $x$ is transcendental.

It is similarly easy to prove that $3^x = 10$ also means that $x$ is transcendental. The leap I'm unable to make is to take that addition sign into account. That addition sign, speaking loosely, gives an additional "degree of freedom" that allows both $6^x$ and $3^x$ to be transcendental (but sum to $10$), which invalidates this line of thinking.

Deepak
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The function $6^x+3^x$ has a minimum at $-\infty $ of zero. and it is an increasing function as well as continuous function so it goes upto infinity at $x=\infty $. So by completeness property of real numbers it has to croos the value of 10 since it lies in between 0 and infinity

Jasser
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