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Let $a>0$ and $b>0$ be real numbers such that $a \ne 1$. Then can

$\exists x. a^x = b$

be proven without using logarithms?

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    Precise answer or estimate answer? – 2017-01-12

1 Answers 1

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Yes: using the fact that the function is continuous, that it diverges on one side and that it tends to zero on the other.

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    Okay, but how do we prove the function is continuous? I suppose that if we want to define $a^x$ without logarithms, we can define it for all rational $x$. It remains to be seen, then, that for a sequence $r_n \to r$ where each $r_n$ is rational (but $r$ might not be), $a^{r_n}$ converges. – 2017-01-12
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    Every function that is convex in the interval $(a,b)$ is continuous for that same interval, and it is easy to see that $f(x) = a^x$ is continuous for the interval $(0,\infty)$. – 2017-01-12
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    I am not saying that the results I use for the proof are trivial. But it is nonetheless true that these can be proven independently of the surjectivity, as asked by the OP. – 2017-01-12
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    Do you mean that $y = a^x$ diverges from 0 to infinity for $a>1$ and converges from infinity to 0 for $a<1$, thus allowing us to use the Intermediate Value Theorem? I was a little confused by the way you put it at first. – 2017-01-12
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    Yes. That's what I am referring to – 2017-01-13