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Let $f(x)$ be a transcendental function with $x\in\mathbb{C}$. Then are the values $f(x)$ themselves transcendental, except perhaps for a "few" exceptions?

For example, it is known that $f(x)=e^x$ is transcendental, and clearly $f(0)=e^0=1$ is algebraic, but what of all the other values? Are they transcendental? What about for other transcendental functions $f$ ?

EDIT: I suppose in the case of $f(x)=e^x$ we have things such as $x=\text{ln}(a),$ where $a$ can be anything we want. What if we only allow algebraic values $x$ ?

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  • $\begingroup$ The function $e^x$ takes on every non-zero rational value. $\endgroup$ – André Nicolas Aug 24 '13 at 22:03
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    $\begingroup$ A nonconstant analytic function on an open subset of $\mathbb C$ has open range, hence its range contains countably infinitely many algebraic numbers. $\endgroup$ – Jonas Meyer Aug 24 '13 at 22:05
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    $\begingroup$ Even if you only allow algebraic values of $x$, remember that $f(x) = 2^x$ is also a transcendental function, but it takes rational values for integer values of $x$ and algebraic values for rational values of $x$. $\endgroup$ – Qiaochu Yuan Aug 24 '13 at 23:15
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    $\begingroup$ All numbers are transcendental except for a few special cases, i.e. all except countably many. As for $e^x$, notice that every positive number belongs to its range. Thus $5$, for example, is a member of the range. $\endgroup$ – Michael Hardy Aug 25 '13 at 2:45
  • $\begingroup$ @MichaelHardy So, for example, $e^{\ln(5)}=5$ or $e^{\ln\sqrt{2}}=\sqrt{2}$, and so on for every algebraic positive number? And it is known from Cantor that the algebraic numbers are countable. That makes sense now! If I understand you correctly I'd accept that as an answer. $\endgroup$ – Pixel Aug 25 '13 at 6:48
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All numbers are transcendental except for a few special cases, i.e. all except countably many. As for $e^x$, notice that every positive number belongs to its range. Thus $5$, for example, is a member of the range.

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There are countably many values of $x$ for which $f(x)$ is algebraic, and even rational. As an example, let $x = \ln{r}$ for any $r \in \mathbb{Q^+}$. In fact, continuous transcendental functions will generally have at least countably many values which are rational, by the intermediate value theorem.

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