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How can I prove that there is no closed form solution to the equation $2^x + 3^x = 10$?

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A great example of why the [algebra-precalculus] vs [abstract-algebra] tag distinction is bogus. If "no closed form" means $x$ is not an algebraic number, then this would follow from Schanuel's conjecture. – T.. Nov 4 '10 at 18:31
A rigorous proof requires a rigorous definition of "closed form solution". – Bill Dubuque Nov 4 '10 at 18:33
Okay. Let say that you can use exponentials, logarithms, digits 0,...,9, variable $x$, $n$th roots, four elementary operations (+ – × ÷) and make compositions and combinations of them. The expression should contain only finitely many characters as written in LaTeX. In particular, symbols $\sum$, $\int$, $\cdots$, $\ldots$ are forbidden. – Jaska Nov 4 '10 at 18:51
@Jaska: in that case you might be interested in reading . – Qiaochu Yuan Nov 4 '10 at 19:04
@Qiaochu Yuan: Thanks for that! – Jaska Nov 4 '10 at 19:15
up vote 1 down vote accepted

There is a solution to this equation, but I don't know that you can make a closed-form solution.

As $\lim_{x\rightarrow -\infty} 2^x + 3^x = 0$ and $\lim_{x \rightarrow \infty} 2^x + 3^x = \infty$ and because $2^x + 3^x$ is continuous there must be a point where

$2^x + 3^x = 10$

by the intermediate value theorem.

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Yes, but this is not the issue at hand. – Qiaochu Yuan Nov 4 '10 at 19:13

By the way, your equation can be written in form


where H is the generalized harmonic number:[3%2C+x]%3D%3D11

So to find x you should investigate the inverse function of generalized harmonic number.

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