Proof that there is no closed form solution to $2^x + 3^x = 10$ How can I prove that there is no closed form solution to this equation? $$2^x + 3^x = 10$$
 A: A closed-form solution is a solution that can be expressed as a closed-form expression.
A mathematical expression is a closed-form expression iff it contains only finite numbers of only constants, explicit functions, operations and/or variables.
Sensefully, all the constants, functions and operations in a given closed-form expression should be from allowed sets.
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The following parts of the answer are only for closed-form solutions that are expressions of elementary functions. According to Liouville and Ritt, the elementary functions can be represented in a finite number of steps by performing only algebraic operations and / or taking exponentials and / or logarithms.
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1)
$2^x+3^x=10$ is a transcendental equation: $e^{\ln(2)x}+e^{\ln(3)x}=10$. The left-hand side of this equation is the functional term of an elementary function. Because  $\ln(2)$ and $\ln(3)$ are linearly independent over $\mathbb{Q}$, the expressions $2^x$ and $3^x$ are algebraically independent: MathStackExchange: Algebraic independence of functions. Therefore one can prove with help of the theorem of [Ritt 1925] (which is also proved in [Risch 1979]) that a function $x\mapsto 2^x+3^x$ cannot have a partial inverse over an open domain $D\subseteq\mathbb{C}$ that is an elementary function. It is not possible therefore to rearrange the equation according to $x$ only by applying only elementary operations (elementary functions) one can read from the equation.
2)
The question of solvability of some special kinds of equations by elementary functions is treated in [Rosenlicht 1969].
3)
$2^x=e^{\ln(2)x}$, $3^x=e^{\ln(3)x}$
$2^x$ and $3^x$ are $\begin{cases}
\text{algebraic}&\text{if }x\text{ is rational}\\
\text{transcendental}&\text{if }x\text{ is algebraic and irrational}\\
&\text{(Gelfond-Schneider theorem)}\\
\text{transcendental}&\text{otherwise(?)}
\end{cases}$
Let $x_0$ be a solution of your equation. If $x_0$ is not rational, $2^{x_0}$ and $3^{x_0}$ are algebraically independent: MathStackExchange: Algebraic independence of functions. That means, $2^{x_0}$ and $3^{x_0}$ cannot fulfill together an algebraic equation, in particular not the equation $2^{x_0}+3^{x_0}=10$. That means, the equation can have only rational solutions.
4)
The existence or non-existence of elementary solutions (that are elementary numbers) could possibly be proved by the methods of [Lin 1983] and [Chow 1999]. But both methods need the Schanuel conjecture that is unproven.
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[Chow 1999] Chow, T. Y.: What is a Closed-Form Number? Amer. Math. Monthly 106 (1999) (5) 440-448 or https://arxiv.org/abs/math/9805045
[Lin 1983] Ferng-Ching Lin: Schanuel's Conjecture Implies Ritt's Conjectures. Chin. J. Math. 11 (1983) (1) 41-50
[Risch 1979] Risch, R. H.: Algebraic Properties of the Elementary Functions of Analysis. Amer. J. Math 101 (1979) (4) 743-759
[Ritt 1925] Ritt, J. F.: Elementary functions and their inverses. Trans. Amer. Math. Soc. 27 (1925) (1) 68-90
[Rosenlicht 1969] Rosenlicht, M.: On the explicit solvability of certain transcendental equations. Publications mathématiques de l'IHÉS 36 (1969) 15-22
A: By the way, your equation can be written in form
$$H^{(-x)}_3=11$$
where H is the generalized harmonic number: https://www.wolframalpha.com/input?i=HarmonicNumber%5B3%2C+-x%5D%3D11
So to find $x$, you should investigate the inverse function of generalized harmonic number.
