How to prove that the logarithm is a transcendental function Let's consider the function $\log x$; how can I prove that it is a transcendental function on the function field of rational functions, i.e. that a polynomial in two variables $p(x,y)$ such that $p(x,\log x)=0$ identically does not exist?
I have been trying different approaches: seeing the logarithm on the real numbers, like a formal series or like a holomorphic function on an open in the complex plane, but I was not able to do this. I have also tried looking for the differences with $\sqrt{(1+x)}$ which is algebraic on the rational function, and can be defined on a subset of the reals, with a formal series or on an open subset of the complex plane.
Could you please help me?
 A: A direct way to do this is to assume there is such a $p(x,\log x)=0, x\in \mathbb R_+$
Since $\log$ takes infinitely many values, $p$ cannot be a polynomial in $\log x$ only, so let $n \ge 1$ the highest power of $x$ with non zero coefficients and write the equation as:
$a_n(\log x)x^n+a_{n-1}(\log x)x^{n-1}+..a_0(\log x)=0$, where $a_0,..a_n$ are polynomials in one variable.
Dividing by $x^n$ and noting that $\frac{a_k(\log x)}{x^{n-k}} \to 0, x \to \infty$ for all $0 \le k \le n-1$, we get that  $a_n(\log x) \to 0, x \to \infty$.
But now if $a_n$ is non constant, obviously $a_n(\log x) \to \pm \infty, x \to \infty$ depending on the sign of its leading coefficient and that is a contradiction. Hence $a_n(\log x)$ is constant and then it must be zero, contradicting the original assumption that the coefficient of $x^n$ in $p$ is non-zero, so we are done!
A: HINT:
Let us show that we cannot have an equality
$$\sum_{k=0}^m P_k(x) \exp ( \lambda_k x) = 0$$
where $P_k(x)$ are non-zero polynomials and $\lambda_k$ are distinct complex numbers. Otherwise we would have
$$P_0(x) \exp (\lambda_0 x) = -\sum_{k=1}^m P_k(x) \exp (\lambda_k x) $$
Now the right hand side is annihilated by a differential operator
$$\prod_{k=1}^m (D - \lambda_k\cdot I)^{d_k+1}$$
while the LHS by $(D-\lambda_0)^{d_0+1}$. But the $\gcd$ of the polynomials $\prod_{k=1}^m(T- \lambda_k)^{d_k+1}$ and $(T-\lambda_0)^{d_0+1}$ is $1$. It follows that both sides are annihilated by the identity differential operator, so both are $0$, contradiction.
