Is ln(n) transcendental for all integer $n > 1$?

Is $\ln(n)$ transcendental for all $n \in \mathbb{N} \setminus \{0, 1\}$? Is the answer even known?

Solution 1:

Yes, $\ln(\alpha)$ is transcendental for any positive algebraic number $\alpha \ne 1$, as a special case of the Lindemann–Weierstrass theorem

Solution 2:

Note that $x = \log{n} \implies e^x = n$ and it is well known that $e^x$ is transcendental if $x$ is algebraic and nonzero, which would give a contradiction if $x$ were algebraic. So $\log{n}$ is transcendental for all $n \geq 2$.

Solution 3:

From the list of known transcendental numbers in the relevant Wikipedia article:

$\ln(a)$ if $a$ is algebraic and not equal to $0$ or $1$

So, in particular, $\ln(1)=0$ is not transcendental (although the numbers $2\pi ik$ for $k\in\mathbb{Z}\setminus\{0\}$ are, and these numbers have just as much right to be called $\ln(1)$; this is what they are referring to when they talk about a "branch of the logarithm function"), but $\ln(n)$ is transcendental for all $n\in\mathbb{N}\setminus\{0,1\}$.

Solution 4:

Lindemann proved $e^α$ is transcendental for every non-zero algebraic number $α$. Let $n$ be a natural number.

Assume that $\log(n)$ is algebraic. It is non-zero if $n>1$. Thus, for $n>1$, we conclude that $e^{\log n}=n$ is transcendental. Contradiction.