Does $6^x+3^x=10$ have an solution that is an algebraic number? 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?
 A: 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.
A: 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
