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It is easy to show $\log$ isn't a polynomial (no continuous extension to $\mathbb{R}$). More challenging is showing it isn't rational.

Suppose it were a rational function. Then write, the fraction in lowest terms

$$\log x =\frac{G(x)}{Q(x)} \Longleftrightarrow\frac{G(x)}{\log x} = Q(x)$$

Clearly, as $x \to 0$, $Q(x) \to 0$. However, $Q(x)$ then has $x$ as factor so that

$$\frac{G(x)}{x\log x} = Q_2(x)$$

It is well known that $x \log x \to 0$ as $x \to 0$, so for $Q_2(x)$ to have a finite limit as $x \to 0$, which it must since it is a polynomial, $G(x) \to 0$ as $x \to 0$ so that $x$ is a factor of $G(x)$. This contradicts the assumption that $\frac{G}{Q}$ was in lowest terms.

If there is something wrong with this, please comment, but my main question is

What are some other ways to prove that $\log x$ isn't a rational function?

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  • $\begingroup$ ($z$ is a complex variable) what you wrote is more or less that rational functions $f(z) = \frac{G(z)}{Q(z)}$ are meromorphic, their singularities always are of the form $f(z) \sim \frac{C}{(z-a)^k}$ as $z \to a$. $\endgroup$ – reuns Jun 10 '16 at 1:51
  • $\begingroup$ Good question +1. It can be proved more generally that $\log x$ is not an algebraic function of $x$. See my answer. $\endgroup$ – Paramanand Singh Jun 10 '16 at 11:24
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One reason is that a rational function is defined over all $\mathbb R$ except for a finite number of points, but $\log$ is not.

Let's prove that $\log$ is not even a rational function restricted to $(0,+\infty)$.

If $\log = \dfrac GQ$, then $\dfrac 1x = \log' = \dfrac{G'Q-GQ'}{Q^2}$.

This implies that $G$ and $Q$ have the same degree and therefore $\displaystyle\lim_{x\to\infty} \dfrac{G(x)}{Q(x)}$ is finite.

But $\displaystyle\lim_{x\to\infty} \log(x) = \infty$.

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  • $\begingroup$ @MathematicsStudent1122: they are defined over all of $\Bbb R$ except finitely many points. $\endgroup$ – Ross Millikan Jun 10 '16 at 2:31
  • $\begingroup$ @RossMillikan Right, but before Ihf edited he simply said they're defined all over $\mathbb{R}$. $\endgroup$ – MathematicsStudent1122 Jun 10 '16 at 2:33
  • $\begingroup$ Just as a small note, the natural logarithm can be extended to the entire complex plane except for zero. These techniques even allow you to solve equations like $\sin(z)=3$. $\endgroup$ – Alex S Oct 19 '17 at 11:25
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More generally we can prove that $\log x$ is not an algebraic function using the following property of $\log x$: $$\lim_{x \to 0^{+}}x^{a}\log x = 0\tag{1}$$ for all positive numbers $a$.

Let's assume on the contrary that $y = \log x$ is an algebraic function. Then we have polynomial functions $a_{1}(x), a_{2}(x), \ldots, a_{n}(x)$ such that $$a_{0}(x)y^{n} + a_{1}(x)y^{n - 1} + \cdots + a_{n - 1}(x)y + a_{n}(x) = 0\tag{2}$$ for all values of $x > 0$. Also note that in the above equation we have both $a_{0}(x), a_{n}(x)$ as non-zero polynomials. Now taking limits for both sides of equation $(2)$ as $x \to 0^{+}$ and using $(1)$ we see that if $b_{i}$ are constant terms of $a_{i}(x)$ then $$\lim_{x \to 0^{+}}(b_{0}y^{n} + b_{1}y^{n - 1} + b_{n - 1}y + \cdots + b_{n}) = 0\tag{3}$$ Dividing by $y$ and noting that $y = \log x \to -\infty$ as $x \to 0^{+}$ we get $$\lim_{x \to 0^{+}}(b_{0}y^{n - 1} + b_{1}y^{n - 2} + \cdots + b_{n - 1}) = 0$$ Repeating the same argument we finally get that $b_{0}, b_{1}, \ldots, b_{n} = 0$. Thus the equation $(2)$ can be divided by $x$ to obtain a similar equation and we can apply the same argument on this new equation. Carrying on this procedure we ultimately get that all the coefficients of the polynomials $a_{i}(x)$ are $0$ and hence all these polynomials are zero polynomials. This contradicts the fact that both $a_{0}(x), a_{n}(x)$ are non-zero polynomials.

For the current question (i.e. prove that $\log x$ is not a rational function) it suffices to take $n = 1$ in the preceding argument.

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A rational function $r$ vanishes at $\infty$ or there is an integer power $q$ so that $r(x)\sim c\cdot x^q$ as $x\to\infty$. The log function exhibits none of these behaviors.

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If $r(x) =\dfrac{a(b)}{b(x)} $, then, as $x \to \infty $, $|r(x)| \sim c|x|^{\deg(a)-\deg(b)} $ for some $c$.

But $\log$ does not go to $\infty$ like this.

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Here's a proof that no antiderivative $F(x)$ of $1/x$ with domain $\Bbb R \setminus \{0\}$ is a rational function.

Such a function is -- as is often forgotten even in calculus textbooks -- of the form $$ F_{C,D}(x) = \begin{cases} \log(-x) + C \quad \, \text{ for } x<0\\ \log(x) +D \qquad\text{ for } x>0\end{cases}$$

with $C,D \in \Bbb R$. It has exactly one vertical asymptote, at $x=0$. So if it were of the form $p(x)/q(x)$ with polynomials $p,q \in \Bbb R[x]$, which wlog we can assume to be without common factor, then $q(x) = x^n$ for some $n \in \Bbb N$. Hence in particular $$x^n \log(x)$$ is a polynomial on $(0,\infty)$. Hence its derivative $$nx^{n-1}\log(x) + x^n \cdot x^{-1} = nx^{n-1}\log(x) + x^{n-1}$$ is a polynomial on $(0,\infty)$ and hence so is $x^{n-1}\log(x)$. Inductively, we get that $\log(x)$ is a polynomial on $(0,\infty)$ which is absurd because of the vertical asymptote at $0$.

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